Limits of Functions: 20 Step-by-Step Practice Problems

\[ \lim_{x\to 2}\left(3x^2-5x+4\right)=6. \]

The function

\[ f(x)=3x^2-5x+4 \]

is a polynomial. Polynomials are defined and continuous at every point of the real line; consequently their limit at a point is obtained by substituting directly for the variable the value which it approaches.

Accordingly we set \(x=2\):

\[ \lim_{x\to 2}\left(3x^2-5x+4\right) =3\cdot 2^2-5\cdot 2+4. \]

Evaluating the individual terms:

\[ 2^2=4,\qquad 3\cdot 4=12,\qquad 5\cdot 2=10. \]

We therefore obtain:

\[ 12-10+4=6. \]

Hence:

\[ \lim_{x\to 2}\left(3x^2-5x+4\right)=6. \]

No indeterminate form arises in this exercise: direct substitution yields at once a well-defined real number.

\[ \lim_{x\to -1}\frac{x^2+3x+5}{2x+5}=1. \]

The limit involves the quotient of two polynomials. Before applying the quotient law for limits we must check that the denominator does not tend to zero.

We evaluate the limit of the denominator by direct substitution:

\[ \lim_{x\to -1}(2x+5)=2(-1)+5=-2+5=3. \]

Since the limit of the denominator is \(3\neq 0\), the quotient law for limits may be applied.

We now evaluate the limit of the numerator:

\[ \lim_{x\to -1}(x^2+3x+5) =(-1)^2+3(-1)+5. \]

Since

\[ (-1)^2=1, \]

we obtain:

\[ 1-3+5=3. \]

Consequently:

\[ \lim_{x\to -1}\frac{x^2+3x+5}{2x+5} = \frac{\displaystyle\lim_{x\to -1}(x^2+3x+5)} {\displaystyle\lim_{x\to -1}(2x+5)} = \frac{3}{3}=1. \]

Hence:

\[\lim_{x\to -1}\frac{x^2+3x+5}{2x+5}=1. \]

Here too direct substitution is sufficient, since the denominator does not vanish in the limit.

\[ \lim_{x\to 3}\frac{x^2-9}{x-3}=6. \]

We begin by substituting \(x=3\) formally in the numerator and in the denominator.

For the numerator we have:

\[ 3^2-9=9-9=0. \]

For the denominator we have:

\[ 3-3=0. \]

Direct substitution therefore leads to the indeterminate form

\[ \frac{0}{0}. \]

This symbol does not represent the value of the limit. It records only that numerator and denominator both tend to zero and that the expression must be transformed.

The numerator is a difference of two squares:

\[ x^2-9=x^2-3^2=(x-3)(x+3). \]

Hence, for \(x\neq 3\), we may write:

\[ \frac{x^2-9}{x-3} = \frac{(x-3)(x+3)}{x-3}. \]

Since in the study of the limit as \(x\to 3\) we consider values of \(x\) close to \(3\) but different from \(3\), the factor \(x-3\) is non-zero and may be cancelled:

\[ \frac{(x-3)(x+3)}{x-3}=x+3. \]

The original limit therefore becomes:

\[ \lim_{x\to 3}\frac{x^2-9}{x-3} = \lim_{x\to 3}(x+3). \]

At this stage direct substitution may be applied:

\[ \lim_{x\to 3}(x+3)=3+3=6. \]

Hence:

\[ \lim_{x\to 3}\frac{x^2-9}{x-3}=6. \]

The cancellation does not alter the limit, because the two expressions

\[ \frac{x^2-9}{x-3} \qquad\text{and}\qquad x+3 \]

agree for every \(x\neq 3\). The fact that the original expression fails to be defined precisely at \(x=3\) is therefore no obstacle to the existence of the limit.

\[ \lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\frac{1}{4}. \]

We first check what direct substitution gives.

The numerator tends to:

\[ \sqrt{4}-2=2-2=0. \]

The denominator tends to:

\[ 4-4=0. \]

We therefore obtain the indeterminate form

\[ \frac{0}{0}. \]

In this case no immediate factorisation of the numerator is available. The most natural transformation is to rationalise, multiplying numerator and denominator by the conjugate of the numerator:

\[ \sqrt{x}+2. \]

For \(x\neq 4\) we may write:

\[ \frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}. \]

The second factor equals \(1\), provided that \(\sqrt{x}+2\neq 0\). On a neighbourhood of \(x=4\) this quantity is positive and hence certainly non-zero.

Multiplying out the numerators produces a difference of two squares:

\[ (\sqrt{x}-2)(\sqrt{x}+2) = (\sqrt{x})^2-2^2 = x-4. \]

Hence:

\[ \frac{\sqrt{x}-2}{x-4} = \frac{x-4}{(x-4)(\sqrt{x}+2)}. \]

For \(x\neq 4\) we may cancel the factor \(x-4\):

\[ \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{\sqrt{x}+2}. \]

The original limit therefore becomes:

\[ \lim_{x\to 4}\frac{\sqrt{x}-2}{x-4} = \lim_{x\to 4}\frac{1}{\sqrt{x}+2}. \]

The denominator now tends to:

\[ \sqrt{4}+2=2+2=4, \]

which is non-zero. Direct substitution may therefore be applied:

\[ \lim_{x\to 4}\frac{1}{\sqrt{x}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{4}. \]

Hence:

\[\lim_{x\to 4}\frac{\sqrt{x}-2}{x-4}=\frac{1}{4}. \]

Rationalisation has removed the indeterminate form by turning the expression into a quotient whose denominator tends to a number different from zero.

\[ \lim_{x\to 1}\frac{x^3-1}{x-1}=3. \]

We begin by examining what direct substitution yields.

For the numerator we have:

\[ 1^3-1=1-1=0. \]

For the denominator we have:

\[ 1-1=0. \]

Direct substitution therefore leads to the indeterminate form

\[ \frac{0}{0}. \]

To remove this indeterminate form, observe that the numerator is a difference of two cubes. Recall the factorisation:

\[ a^3-b^3=(a-b)(a^2+ab+b^2). \]

Setting \(a=x\) and \(b=1\), we obtain:

\[ x^3-1=(x-1)(x^2+x+1). \]

For \(x\neq 1\) the quotient may therefore be rewritten in the form:

\[ \frac{x^3-1}{x-1} = \frac{(x-1)(x^2+x+1)}{x-1}. \]

Since \(x\neq 1\), the factor \(x-1\) is non-zero and may be cancelled:

\[ \frac{(x-1)(x^2+x+1)}{x-1} = x^2+x+1. \]

The original limit thus becomes:

\[ \lim_{x\to 1}\frac{x^3-1}{x-1} = \lim_{x\to 1}(x^2+x+1). \]

The resulting expression is a polynomial, so that the limit is evaluated by direct substitution:

\[ \lim_{x\to 1}(x^2+x+1) = 1^2+1+1 = 3. \]

Hence:

\[ \lim_{x\to 1}\frac{x^3-1}{x-1}=3. \]

Here too the cancellation is legitimate because, in computing the limit, we consider values of \(x\) close to \(1\) but different from \(1\). The two expressions therefore agree on a punctured neighbourhood of the point under consideration.

\[ \lim_{x\to 9}\frac{x-9}{\sqrt{x}-3}=6. \]

On substituting \(x=9\) formally, the numerator tends to:

\[ 9-9=0, \]

whereas the denominator tends to:

\[ \sqrt{9}-3=3-3=0. \]

We are therefore faced with the indeterminate form

\[ \frac{0}{0}. \]

In this exercise the radical expression occurs in the denominator. To remove the indeterminate form, we multiply numerator and denominator by the conjugate of the denominator:

\[ \sqrt{x}+3. \]

For \(x\neq 9\) we may write:

\[ \frac{x-9}{\sqrt{x}-3} \cdot \frac{\sqrt{x}+3}{\sqrt{x}+3}. \]

Multiplying out the denominators produces a difference of two squares:

\[ (\sqrt{x}-3)(\sqrt{x}+3) = (\sqrt{x})^2-3^2 = x-9. \]

Hence:

\[ \frac{x-9}{\sqrt{x}-3} = \frac{(x-9)(\sqrt{x}+3)}{x-9}. \]

For \(x\neq 9\) the factor \(x-9\) is non-zero and may be cancelled:

\[ \frac{(x-9)(\sqrt{x}+3)}{x-9} = \sqrt{x}+3. \]

The original limit therefore becomes:

\[ \lim_{x\to 9}\frac{x-9}{\sqrt{x}-3} = \lim_{x\to 9}(\sqrt{x}+3). \]

Direct substitution may now be applied:

\[ \lim_{x\to 9}(\sqrt{x}+3) = \sqrt{9}+3 = 3+3 = 6. \]

Hence:

\[ \lim_{x\to 9}\frac{x-9}{\sqrt{x}-3}=6. \]

Rationalisation has replaced the original quotient by an expression equal to it for \(x\neq 9\), for which direct substitution no longer produces an indeterminate form.

\[ \lim_{x\to+\infty}\frac{4x^2-3x+1}{2x^2+x-5}=2. \]

As \(x\to+\infty\), both the numerator and the denominator become arbitrarily large. Formal substitution therefore leads to the indeterminate form

\[ \frac{\infty}{\infty}. \]

This form does not by itself determine the limit. We must compare the growth of the terms occurring in the numerator and in the denominator.

The term of highest degree is \(x^2\), both in the numerator and in the denominator. We therefore divide every term by \(x^2\):

\[ \frac{4x^2-3x+1}{2x^2+x-5} = \frac{\displaystyle\frac{4x^2}{x^2} -\displaystyle\frac{3x}{x^2} +\displaystyle\frac{1}{x^2}} {\displaystyle\frac{2x^2}{x^2} +\displaystyle\frac{x}{x^2} -\displaystyle\frac{5}{x^2}}. \]

Simplifying each quotient, we obtain:

\[ \frac{4x^2-3x+1}{2x^2+x-5} = \frac{4-\displaystyle\frac{3}{x}+\displaystyle\frac{1}{x^2}} {2+\displaystyle\frac{1}{x}-\displaystyle\frac{5}{x^2}}. \]

As \(x\to+\infty\), the following limits hold:

\[ \frac{1}{x}\to 0 \qquad\text{and}\qquad \frac{1}{x^2}\to 0. \]

Consequently the numerator tends to:

\[ 4-3\cdot 0+0=4, \]

while the denominator tends to:

\[ 2+0-5\cdot 0=2. \]

Since the limit of the denominator is non-zero, the quotient law for limits may be applied:

\[ \lim_{x\to+\infty} \frac{4-\displaystyle\frac{3}{x}+\displaystyle\frac{1}{x^2}} {2+\displaystyle\frac{1}{x}-\displaystyle\frac{5}{x^2}} = \frac{4}{2} = 2. \]

Hence:

\[ \lim_{x\to+\infty}\frac{4x^2-3x+1}{2x^2+x-5}=2. \]

The result agrees with the ratio of the coefficients of the terms of highest degree:

\[ \frac{4}{2}=2. \]

This happens because, for very large values of \(x\), the terms of lower degree become negligible in comparison with the quadratic terms.

\[ \lim_{x\to-\infty}\frac{3x^2+2}{x^3-1}=0. \]

As \(x\to-\infty\), the numerator \(3x^2+2\) tends to \(+\infty\), since \(x^2\) is always non-negative and becomes arbitrarily large.

The denominator \(x^3-1\), on the other hand, tends to \(-\infty\), since a power with odd exponent preserves the sign of \(x\).

Formal substitution therefore leads to a quotient of infinitely large quantities. To determine the limit we divide numerator and denominator by the highest power occurring in the fraction as a whole, namely \(x^3\):

\[ \frac{3x^2+2}{x^3-1} = \frac{\displaystyle\frac{3x^2}{x^3} +\displaystyle\frac{2}{x^3}} {\displaystyle\frac{x^3}{x^3} -\displaystyle\frac{1}{x^3}}. \]

Simplifying the individual terms, we obtain:

\[ \frac{3x^2+2}{x^3-1} = \frac{\displaystyle\frac{3}{x} +\displaystyle\frac{2}{x^3}} {1-\displaystyle\frac{1}{x^3}}. \]

As \(x\to-\infty\), we have:

\[ \frac{1}{x}\to 0, \qquad \frac{1}{x^3}\to 0. \]

Consequently the numerator tends to:

\[ 3\cdot 0+2\cdot 0=0, \]

while the denominator tends to:

\[ 1-0=1. \]

Since the limit of the denominator is non-zero, the quotient law for limits may be applied:

\[ \lim_{x\to-\infty} \frac{\displaystyle\frac{3}{x} +\displaystyle\frac{2}{x^3}} {1-\displaystyle\frac{1}{x^3}} = \frac{0}{1} = 0. \]

Hence:

\[ \lim_{x\to-\infty}\frac{3x^2+2}{x^3-1}=0. \]

The denominator has higher degree than the numerator. It therefore grows in absolute value more rapidly than the numerator, and the quotient tends to zero.

In particular, for negative values of \(x\) sufficiently large in absolute value, the numerator is positive and the denominator is negative. The quotient thus approaches zero through negative values, but the value of the limit is simply \(0\).

\[ \lim_{x\to-\infty}\frac{2x^3-x}{x^2+4}=-\infty. \]

As \(x\to-\infty\), the dominant term of the numerator is \(2x^3\). Since \(x^3\to-\infty\), we have:

\[ 2x^3-x\to-\infty. \]

The dominant term of the denominator is \(x^2\). Since \(x^2\to+\infty\), we have:

\[ x^2+4\to+\infty. \]

We therefore obtain an expression of the type

\[ \frac{-\infty}{+\infty}, \]

which must be examined by comparing the orders of growth of the numerator and of the denominator.

The numerator has degree \(3\), whereas the denominator has degree \(2\). The numerator therefore grows in absolute value more rapidly than the denominator. To make this behaviour explicit, we divide numerator and denominator by \(x^2\):

\[ \frac{2x^3-x}{x^2+4} = \frac{\displaystyle\frac{2x^3}{x^2} -\displaystyle\frac{x}{x^2}} {\displaystyle\frac{x^2}{x^2} +\displaystyle\frac{4}{x^2}}. \]

Simplifying, we obtain:

\[ \frac{2x^3-x}{x^2+4} = \frac{2x-\displaystyle\frac{1}{x}} {1+\displaystyle\frac{4}{x^2}}. \]

As \(x\to-\infty\), we have:

\[ 2x\to-\infty, \qquad \frac{1}{x}\to 0, \qquad \frac{4}{x^2}\to 0. \]

Consequently the numerator

\[ 2x-\frac{1}{x} \]

tends to \(-\infty\), while the denominator

\[ 1+\frac{4}{x^2} \]

tends to \(1\).

Moreover, the denominator is always positive, since

\[ 1+\frac{4}{x^2}>0 \]

for every \(x\neq 0\). Dividing a quantity which tends to \(-\infty\) by a positive quantity which tends to \(1\) therefore does not affect the sign of the divergence.

Hence:

\[ \lim_{x\to-\infty}\frac{2x^3-x}{x^2+4}=-\infty. \]

The result may also be read off by observing that the quotient behaves asymptotically like

\[ \frac{2x^3}{x^2}=2x, \]

and \(2x\to-\infty\) as \(x\to-\infty\).

\[ \lim_{x\to+\infty}\left(\sqrt{x^2+4x}-x\right)=2. \]

As \(x\to+\infty\), we have:

\[ \sqrt{x^2+4x}\to+\infty \qquad\text{and}\qquad x\to+\infty. \]

Formal substitution therefore leads to the indeterminate form

\[ \infty-\infty. \]

We cannot conclude that the limit is zero, since the difference of two divergent quantities depends on the rate at which they grow. The expression must be transformed.

We rationalise by multiplying and dividing by the conjugate:

\[ \sqrt{x^2+4x}+x. \]

This gives:

\[ \sqrt{x^2+4x}-x = \frac{(\sqrt{x^2+4x}-x)(\sqrt{x^2+4x}+x)} {\sqrt{x^2+4x}+x}. \]

A difference of two squares appears in the numerator:

\[ (\sqrt{x^2+4x})^2-x^2 = x^2+4x-x^2 = 4x. \]

Hence:

\[ \sqrt{x^2+4x}-x = \frac{4x}{\sqrt{x^2+4x}+x}. \]

To study the limit we divide numerator and denominator by \(x\). Since \(x\to+\infty\), in particular \(x>0\) for sufficiently large values, and therefore

\[ \sqrt{x^2}=|x|=x. \]

We may then write:

\[ \frac{4x}{\sqrt{x^2+4x}+x} = \frac{4} {\displaystyle\frac{\sqrt{x^2+4x}}{x}+1}. \]

Inside the radical, we factor out \(x^2\):

\[ \sqrt{x^2+4x} = \sqrt{x^2\left(1+\frac{4}{x}\right)} = x\sqrt{1+\frac{4}{x}}. \]

Consequently:

\[ \frac{\sqrt{x^2+4x}}{x} = \sqrt{1+\frac{4}{x}}. \]

The expression thus becomes:

\[ \frac{4} {\sqrt{1+\displaystyle\frac{4}{x}}+1}. \]

Since

\[ \frac{4}{x}\to 0, \]

we have:

\[ \sqrt{1+\frac{4}{x}}\to 1. \]

The denominator therefore tends to:

\[ 1+1=2. \]

Hence:

\[ \lim_{x\to+\infty} \frac{4} {\sqrt{1+\displaystyle\frac{4}{x}}+1} = \frac{4}{2} = 2. \]

We conclude that:

\[ \lim_{x\to+\infty}\left(\sqrt{x^2+4x}-x\right)=2. \]

Rationalisation has converted the form \(\infty-\infty\) into a quotient whose limit is determinate.

\[ \lim_{x\to 0}\frac{\sin(3x)}{x}=3. \]

On substituting \(x=0\) formally, the numerator tends to:

\[ \sin(3\cdot 0)=\sin 0=0, \]

while the denominator tends to zero. We are therefore faced with the indeterminate form

\[ \frac{0}{0}. \]

To evaluate the limit we use the standard limit:

\[ \lim_{t\to 0}\frac{\sin t}{t}=1. \]

In the given expression the argument of the sine is not \(x\) but \(3x\). We must therefore make \(3x\) appear in the denominator as well.

We multiply and divide by \(3\):

\[ \frac{\sin(3x)}{x} = 3\frac{\sin(3x)}{3x}. \]

This identity holds for \(x\neq 0\), a condition consistent with the computation of the limit as \(x\to 0\).

As \(x\to 0\), we also have \(3x\to 0\). The standard limit may therefore be applied on setting

\[ t=3x. \]

This gives:

\[ \lim_{x\to 0}\frac{\sin(3x)}{3x}=1. \]

Consequently:

\[ \lim_{x\to 0}\frac{\sin(3x)}{x} = 3\lim_{x\to 0}\frac{\sin(3x)}{3x} = 3\cdot 1 = 3. \]

Hence:

\[ \lim_{x\to 0}\frac{\sin(3x)}{x}=3. \]

In terms of asymptotic equivalence, since

\[ \sin(3x)\sim 3x \qquad\text{as }x\to 0, \]

the quotient behaves like

\[ \frac{3x}{x}=3. \]

\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}. \]

On substituting \(x=0\) formally, the numerator tends to:

\[ 1-\cos 0=1-1=0, \]

while the denominator tends to:

\[ 0^2=0. \]

We are therefore faced with the indeterminate form

\[ \frac{0}{0}. \]

The standard limit may be applied directly:

\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}. \]

To see where this result comes from, it may be reduced to the standard limit for the sine by means of the trigonometric identity:

\[ 1-\cos x=2\sin^2\left(\frac{x}{2}\right). \]

Substituting this identity into the quotient, we obtain:

\[ \frac{1-\cos x}{x^2} = \frac{2\sin^2\left(\displaystyle\frac{x}{2}\right)}{x^2}. \]

In order to bring out the standard quotient, observe that:

\[ x^2=4\left(\frac{x}{2}\right)^2. \]

Hence:

\[ \frac{2\sin^2\left(\displaystyle\frac{x}{2}\right)}{x^2} = \frac{2\sin^2\left(\displaystyle\frac{x}{2}\right)} {4\left(\displaystyle\frac{x}{2}\right)^2}. \]

Cancelling the numerical factor:

\[ \frac{1-\cos x}{x^2} = \frac{1}{2} \frac{\sin^2\left(\displaystyle\frac{x}{2}\right)} {\left(\displaystyle\frac{x}{2}\right)^2}. \]

The second factor may be written as the square of a quotient:

\[ \frac{\sin^2\left(\displaystyle\frac{x}{2}\right)} {\left(\displaystyle\frac{x}{2}\right)^2} = \left( \frac{\sin\left(\displaystyle\frac{x}{2}\right)} {\displaystyle\frac{x}{2}} \right)^2. \]

Consequently:

\[ \frac{1-\cos x}{x^2} = \frac{1}{2} \left( \frac{\sin\left(\displaystyle\frac{x}{2}\right)} {\displaystyle\frac{x}{2}} \right)^2. \]

As \(x\to 0\), we also have \(\displaystyle\frac{x}{2}\to 0\). By the standard limit for the sine:

\[ \lim_{x\to 0} \frac{\sin\left(\displaystyle\frac{x}{2}\right)} {\displaystyle\frac{x}{2}} = 1. \]

Hence:

\[ \lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2}\cdot 1^2 = \frac{1}{2}. \]

We conclude that:

\[ \lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}. \]

The same result may be read off from the asymptotic equivalence

\[ 1-\cos x\sim\frac{x^2}{2} \qquad\text{as }x\to 0. \]

\[ \lim_{x\to 0}\frac{e^{2x}-1}{x}=2. \]

On substituting \(x=0\) formally, the numerator tends to:

\[ e^{2\cdot 0}-1=e^0-1=1-1=0, \]

while the denominator tends to zero. We are therefore faced with the indeterminate form

\[ \frac{0}{0}. \]

To evaluate the limit we use the standard limit:

\[ \lim_{t\to 0}\frac{e^t-1}{t}=1. \]

In the given expression the exponent is \(2x\). In order to apply the standard limit we must therefore make \(2x\) appear in the denominator as well.

For \(x\neq 0\) we may write:

\[ \frac{e^{2x}-1}{x} = 2\frac{e^{2x}-1}{2x}. \]

As \(x\to 0\), we also have \(2x\to 0\). On setting

\[ t=2x, \]

the standard limit gives:

\[ \lim_{x\to 0}\frac{e^{2x}-1}{2x}=1. \]

Hence:

\[ \lim_{x\to 0}\frac{e^{2x}-1}{x} = 2\lim_{x\to 0}\frac{e^{2x}-1}{2x} = 2\cdot 1 = 2. \]

We conclude that:

\[ \lim_{x\to 0}\frac{e^{2x}-1}{x}=2. \]

The same result may be obtained from the asymptotic equivalence

\[ e^{2x}-1\sim 2x \qquad\text{as }x\to 0. \]

Consequently the numerator of the quotient may be replaced by its equivalent:

\[ \frac{e^{2x}-1}{x} \sim \frac{2x}{x} = 2. \]

\[ \lim_{x\to 0}\frac{\ln(1+5x)}{x}=5. \]

Before proceeding, note that the function is defined when

\[ 1+5x>0. \]

This condition is equivalent to:

\[ x>-\frac{1}{5}. \]

In particular, the function is defined on a whole neighbourhood of \(x=0\), so that the two-sided limit is meaningful.

On substituting \(x=0\) formally, the numerator tends to:

\[ \ln(1+5\cdot 0)=\ln(1)=0, \]

while the denominator tends to zero. We therefore have the indeterminate form

\[ \frac{0}{0}. \]

We use the standard limit:

\[ \lim_{t\to 0}\frac{\ln(1+t)}{t}=1. \]

In the given expression the quantity added to \(1\) is \(5x\). In order to apply the standard limit, we make \(5x\) appear in the denominator as well:

\[ \frac{\ln(1+5x)}{x} = 5\frac{\ln(1+5x)}{5x}. \]

As \(x\to 0\), we also have \(5x\to 0\). On setting

\[ t=5x, \]

we obtain:

\[ \lim_{x\to 0}\frac{\ln(1+5x)}{5x}=1. \]

Hence:

\[ \lim_{x\to 0}\frac{\ln(1+5x)}{x} = 5\lim_{x\to 0}\frac{\ln(1+5x)}{5x} = 5\cdot 1 = 5. \]

We conclude that:

\[ \lim_{x\to 0}\frac{\ln(1+5x)}{x}=5. \]

In terms of asymptotic equivalence, since

\[ \ln(1+5x)\sim 5x \qquad\text{as }x\to 0, \]

the quotient behaves like:

\[ \frac{5x}{x}=5. \]

\[ \lim_{x\to 0} \frac{\bigl(1-\cos x\bigr)\ln(1+2x)} {x\sin(3x)} =0. \]

On substituting \(x=0\) formally, every factor in the numerator and in the denominator tends to zero:

\[ 1-\cos x\to 0, \qquad \ln(1+2x)\to 0, \qquad x\to 0, \qquad \sin(3x)\to 0. \]

The quotient therefore takes the indeterminate form

\[ \frac{0}{0}. \]

In this case it is convenient to use the basic asymptotic equivalences:

\[ 1-\cos x\sim\frac{x^2}{2}, \qquad \ln(1+2x)\sim 2x, \qquad \sin(3x)\sim 3x, \]

valid as \(x\to 0\).

Such equivalences may be substituted directly inside products and quotients. The given quotient is therefore equivalent to:

\[ \frac{\displaystyle\frac{x^2}{2}\cdot 2x} {x\cdot 3x}. \]

Simplifying the numerator:

\[ \frac{x^2}{2}\cdot 2x=x^3. \]

Simplifying the denominator as well:

\[ x\cdot 3x=3x^2. \]

Hence:

\[ \frac{\bigl(1-\cos x\bigr)\ln(1+2x)} {x\sin(3x)} \sim \frac{x^3}{3x^2} = \frac{x}{3}. \]

Since

\[ \lim_{x\to 0}\frac{x}{3}=0, \]

it follows that the given limit is likewise equal to zero:

\[ \lim_{x\to 0} \frac{\bigl(1-\cos x\bigr)\ln(1+2x)} {x\sin(3x)} =0. \]

It is important to observe that the equivalences have been used inside products and quotients, where replacement by an equivalent function is legitimate.

It would not, on the other hand, be legitimate to replace a term automatically by an equivalent one inside a sum or a difference, since in such situations cancellation between the leading terms may occur.

\[ \lim_{x\to 0}x^2\cos\left(\frac{1}{x}\right)=0. \]

The function

\[ \cos\left(\frac{1}{x}\right) \]

has no limit as \(x\to 0\). Indeed, as \(x\) approaches zero, the quantity \(\displaystyle\frac{1}{x}\) becomes unbounded in absolute value, while the cosine continues to oscillate between \(-1\) and \(1\).

The product law for limits cannot therefore be applied directly, since one of the two factors has no limit.

We may, however, use the fact that, for every real number \(t\),

\[ -1\leq\cos t\leq 1. \]

On setting

\[ t=\frac{1}{x}, \]

we obtain, for every \(x\neq 0\):

\[ -1\leq\cos\left(\frac{1}{x}\right)\leq 1. \]

We multiply each member by \(x^2\). Since

\[ x^2\geq 0, \]

the direction of the inequalities is unchanged:

\[ -x^2 \leq x^2\cos\left(\frac{1}{x}\right) \leq x^2. \]

We now evaluate the limits of the two functions which bound the expression:

\[ \lim_{x\to 0}(-x^2)=0 \qquad\text{and}\qquad \lim_{x\to 0}x^2=0. \]

The middle function therefore lies between two functions which both tend to the same limit \(0\).

By the squeeze theorem it follows that:

\[ \lim_{x\to 0}x^2\cos\left(\frac{1}{x}\right)=0. \]

The oscillating factor has no limit, but remains bounded between \(-1\) and \(1\). The factor \(x^2\), on the other hand, tends to zero and makes the amplitude of the oscillations of the product ever smaller.

The same argument may also be expressed by means of the absolute value:

\[ \left| x^2\cos\left(\frac{1}{x}\right) \right| = x^2\left| \cos\left(\frac{1}{x}\right) \right| \leq x^2. \]

Since \(x^2\to 0\), the absolute value of the expression tends to zero, and hence so does the expression itself.

\[ \lim_{x\to 2^-}\frac{1}{x-2}=-\infty, \qquad \lim_{x\to 2^+}\frac{1}{x-2}=+\infty. \]

Since the two one-sided limits are different, the limit

\[ \lim_{x\to 2}\frac{1}{x-2} \]

does not exist.

The function

\[ f(x)=\frac{1}{x-2} \]

is not defined at \(x=2\), since the denominator vanishes there.

In order to study the behaviour of the function near \(2\), it is not enough to observe that the denominator tends to zero. The sign with which it approaches zero must also be determined.

Consider first the left-hand limit:

\[ \lim_{x\to 2^-}\frac{1}{x-2}. \]

The notation \(x\to 2^-\) means that \(x\) takes values less than \(2\) and ever closer to \(2\). Consequently:

\[ x-2<0. \]

Moreover, as \(x\) approaches \(2\) from the left, the difference \(x-2\) approaches zero through negative values. This is written concisely as:

\[ x-2\to 0^-. \]

The reciprocal of a negative number very close to zero is a negative number very large in absolute value. For example:

\[ \frac{1}{-0.1}=-10, \qquad \frac{1}{-0.01}=-100, \qquad \frac{1}{-0.001}=-1000. \]

Hence:

\[ \lim_{x\to 2^-}\frac{1}{x-2}=-\infty. \]

Now consider the right-hand limit:

\[ \lim_{x\to 2^+}\frac{1}{x-2}. \]

The notation \(x\to 2^+\) means that \(x\) takes values greater than \(2\) and ever closer to \(2\). Consequently:

\[ x-2>0. \]

As \(x\) approaches \(2\) from the right, the difference \(x-2\) approaches zero through positive values:

\[ x-2\to 0^+. \]

The reciprocal of a positive number very close to zero is an arbitrarily large positive number. For example:

\[ \frac{1}{0.1}=10, \qquad \frac{1}{0.01}=100, \qquad \frac{1}{0.001}=1000. \]

Hence:

\[ \lim_{x\to 2^+}\frac{1}{x-2}=+\infty. \]

We have therefore obtained:

\[ \lim_{x\to 2^-}\frac{1}{x-2}=-\infty \qquad\text{and}\qquad \lim_{x\to 2^+}\frac{1}{x-2}=+\infty. \]

For the two-sided limit to exist, the two one-sided limits must exist and be equal. Here they have opposite signs, and therefore:

\[ \lim_{x\to 2}\frac{1}{x-2}\text{ does not exist}. \]

Since at least one of the one-sided limits is infinite, the line

\[ x=2 \]

is a vertical asymptote of the graph of the function.

\[ \lim_{x\to 0^-}\frac{x^2+x}{|x|}=-1, \qquad \lim_{x\to 0^+}\frac{x^2+x}{|x|}=1. \]

Since the two one-sided limits are different,

\[ \lim_{x\to 0}\frac{x^2+x}{|x|} \]

does not exist.

On substituting \(x=0\) formally, the numerator tends to:

\[ 0^2+0=0, \]

while the denominator tends to:

\[ |0|=0. \]

We are therefore faced with the indeterminate form

\[ \frac{0}{0}. \]

The presence of the absolute value makes it necessary to distinguish the behaviour of the function to the right and to the left of \(0\).

Recall indeed that:

\[ |x|= \begin{cases} x, & x\geq 0,\\ -x, & x<0. \end{cases} \]

We begin with the right-hand limit. If \(x\to 0^+\), then \(x>0\) and hence:

\[ |x|=x. \]

For \(x>0\) the expression becomes:

\[ \frac{x^2+x}{|x|} = \frac{x^2+x}{x}. \]

We take out a factor \(x\) in the numerator:

\[ x^2+x=x(x+1). \]

Since \(x\neq 0\), we may cancel:

\[ \frac{x(x+1)}{x}=x+1. \]

Hence:

\[ \lim_{x\to 0^+}\frac{x^2+x}{|x|} = \lim_{x\to 0^+}(x+1) = 1. \]

Now consider the left-hand limit. If \(x\to 0^-\), then \(x<0\) and hence:

\[ |x|=-x. \]

For \(x<0\) the expression becomes:

\[ \frac{x^2+x}{|x|} = \frac{x^2+x}{-x}. \]

Taking out a factor \(x\) in the numerator, we obtain:

\[ \frac{x(x+1)}{-x}. \]

Since \(x\neq 0\), we may cancel the factor \(x\):

\[ \frac{x(x+1)}{-x}=-(x+1). \]

Hence:

\[ \lim_{x\to 0^-}\frac{x^2+x}{|x|} = \lim_{x\to 0^-}\bigl(-(x+1)\bigr) = -1. \]

The two one-sided limits are therefore:

\[ \lim_{x\to 0^-}\frac{x^2+x}{|x|}=-1, \qquad \lim_{x\to 0^+}\frac{x^2+x}{|x|}=1. \]

Since they are different, the two-sided limit does not exist:

\[ \lim_{x\to 0}\frac{x^2+x}{|x|}\text{ does not exist}. \]

The exercise shows that, in the presence of an absolute value, one and the same expression may take different algebraic forms on the two sides of the point under consideration.

\[ \lim_{x\to 0}(1+2x)^{\frac{3}{x}}=e^6. \]

We first check that the power is defined on a neighbourhood of \(x=0\). The base must be positive:

\[ 1+2x>0. \]

This condition is equivalent to:

\[ x>-\frac{1}{2}. \]

The base is therefore positive on a whole neighbourhood of \(0\), and the two-sided limit is meaningful.

As \(x\to 0\), the base tends to:

\[ 1+2x\to 1. \]

The exponent, however, behaves differently on the two sides:

\[ \frac{3}{x}\to+\infty \quad\text{as }x\to0^+, \qquad \frac{3}{x}\to-\infty \quad\text{as }x\to0^-. \]

We are therefore faced with an indeterminate exponential form: \(1^{+\infty}\) for the right-hand limit and \(1^{-\infty}\) for the left-hand limit. In both cases one must study the product of the exponent with the natural logarithm of the base.

To do so, define:

\[ y(x)=(1+2x)^{\frac{3}{x}}. \]

Since \(1+2x>0\) on a neighbourhood of \(0\), we have \(y(x)>0\), and hence we may take natural logarithms:

\[ \ln\bigl(y(x)\bigr)=\ln\left((1+2x)^{\frac{3}{x}}\right). \]

Using the identity

\[ \ln(a^b)=b\ln(a), \qquad a>0, \]

we obtain:

\[ \ln\bigl(y(x)\bigr)=\frac{3}{x}\ln(1+2x). \]

We rewrite the right-hand side so as to bring out the standard logarithmic limit:

\[ \frac{3}{x}\ln(1+2x)=6\frac{\ln(1+2x)}{2x}. \]

As \(x\to 0\), we also have \(2x\to 0\). By the standard limit

\[ \lim_{t\to 0}\frac{\ln(1+t)}{t}=1, \]

it follows that:

\[ \lim_{x\to 0}\frac{\ln(1+2x)}{2x}=1. \]

Hence:

\[ \lim_{x\to 0}\ln\bigl(y(x)\bigr)=6. \]

Since

\[ y(x)=e^{\ln(y(x))}, \]

the continuity of the exponential function gives:

\[ \lim_{x\to 0}y(x)=e^{\displaystyle\lim_{x\to 0}\ln(y(x))}=e^6. \]

We conclude that:

\[ \lim_{x\to 0}(1+2x)^{\frac{3}{x}}=e^6. \]

The result may also be recognised by reducing the expression to the fundamental limit

\[ \lim_{t\to 0}(1+t)^{1/t}=e, \]

but the method by way of the logarithm makes every step explicit and applies systematically to indeterminate exponential forms.

The function has

\[ x=1 \]

as a vertical asymptote and

\[ y=2x+5 \]

as an oblique asymptote, both as \(x\to+\infty\) and as \(x\to-\infty\).

There are no horizontal asymptotes.

The function is a rational function:

\[ f(x)=\frac{2x^2+3x-1}{x-1}. \]

The denominator vanishes when:

\[ x-1=0, \]

that is, for:

\[ x=1. \]

The domain of the function is therefore:

\[ \mathbb{R}\setminus\{1\}. \]

In order to decide whether the line \(x=1\) is a vertical asymptote, we must study the behaviour of the function as \(x\) tends to \(1\).

We compute the value approached by the numerator:

\[ 2\cdot 1^2+3\cdot 1-1 = 2+3-1 = 4. \]

The numerator therefore tends to a positive number different from zero, while the denominator tends to zero.

We consider the two sides separately.

If \(x\to 1^-\), then:

\[ x-1\to 0^-. \]

The numerator remains positive for \(x\) sufficiently close to \(1\), since it tends to the positive number \(4\). The quotient of a positive quantity close to \(4\) by a negative quantity close to zero therefore tends to \(-\infty\):

\[ \lim_{x\to 1^-}\frac{2x^2+3x-1}{x-1} = -\infty. \]

If, on the other hand, \(x\to 1^+\), then:

\[ x-1\to 0^+. \]

The quotient of a positive quantity close to \(4\) by a positive quantity close to zero tends to \(+\infty\):

\[ \lim_{x\to 1^+}\frac{2x^2+3x-1}{x-1} = +\infty. \]

Since at least one of the two one-sided limits is infinite, the line

\[x=1 \]

is a vertical asymptote.

We now study the behaviour of the function as \(x\to+\infty\) and as \(x\to-\infty\).

The numerator has degree \(2\), whereas the denominator has degree \(1\). Since the degree of the numerator exceeds that of the denominator by one, an oblique asymptote may exist.

We carry out the division of the two polynomials:

\[ 2x^2+3x-1=(x-1)(2x+5)+4. \]

Let us check the product:

\[ (x-1)(2x+5) = 2x^2+5x-2x-5 = 2x^2+3x-5. \]

Adding the remainder \(4\), we obtain:

\[ 2x^2+3x-5+4 = 2x^2+3x-1, \]

as required.

Dividing both members by \(x-1\), for \(x\neq 1\), we deduce:

\[ \frac{2x^2+3x-1}{x-1} = 2x+5+\frac{4}{x-1}. \]

In order to verify that the line

\[ y=2x+5 \]

is an oblique asymptote, we compute the difference between the function and the equation of the line:

\[ f(x)-(2x+5) = \frac{4}{x-1}. \]

As \(x\to+\infty\), we have:

\[ \lim_{x\to+\infty}\frac{4}{x-1}=0. \]

Hence:

\[ \lim_{x\to+\infty}\bigl[f(x)-(2x+5)\bigr]=0. \]

As \(x\to-\infty\) we likewise have:

\[ \lim_{x\to-\infty}\frac{4}{x-1}=0, \]

and therefore:

\[ \lim_{x\to-\infty}\bigl[f(x)-(2x+5)\bigr]=0. \]

Since the difference \(f(x)-(2x+5)\) tends to zero in both directions, the graph approaches the line \(y=2x+5\) as \(x\to\pm\infty\).

Hence:

\[ y=2x+5 \]

is an oblique asymptote both as \(x\to+\infty\) and as \(x\to-\infty\).

The function has no horizontal asymptotes. Indeed, as \(x\to\pm\infty\) the dominant term is \(2x\), so that the function does not tend to a finite real number.

In conclusion, the asymptotes of the function are:

\[ x=1 \qquad\text{and}\qquad y=2x+5. \]