The following exercises apply the definitions and properties of infinite limits and limits at infinity. They cover infinite limits at a point, one-sided limits, vertical and horizontal asymptotes, indeterminate forms, comparisons between functions, and the identification of the dominant term.
Each exercise is worked through step by step, with particular attention paid to the sign of quantities tending to zero, distinguishing between left-hand and right-hand limits whenever this distinction is necessary.
For limits at infinity we shall use only the tools introduced in the theory: the properties of limits, algebraic manipulation, rationalisation, comparison of degrees, identification of the dominant term, and the comparison theorem.
Exercise 1 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to2}\frac{1}{(x-2)^2} \]
and verify it directly from the definition of a limit equal to \(+\infty\).
Answer
\[ \lim_{x\to2}\frac{1}{(x-2)^2}=+\infty. \]
Solution
As \(x\to2\), the difference \(x-2\) tends to \(0\). Since this difference is squared, for every \(x\neq2\) we have
\[ (x-2)^2>0. \]
The denominator therefore tends to zero through positive values:
\[ (x-2)^2\to0^+. \]
The reciprocal of a positive quantity tending to zero grows without bound. We therefore expect
\[ \lim_{x\to2}\frac{1}{(x-2)^2}=+\infty. \]
Let us now verify this result directly from the definition. We must show that, for every \(M>0\), there exists \(\delta>0\) such that
\[ 0<|x-2|<\delta \]
implies
\[ \frac{1}{(x-2)^2}>M. \]
We start from the inequality we wish to obtain. Since \((x-2)^2>0\) and \(M>0\),
\[ \frac{1}{(x-2)^2}>M \]
is equivalent to
\[ (x-2)^2<\frac{1}{M}. \]
Taking square roots,
\[ |x-2|<\frac{1}{\sqrt{M}}. \]
It therefore suffices to choose
\[ \delta=\frac{1}{\sqrt{M}}. \]
Indeed, if \(0<|x-2|<\delta\), then
\[ |x-2|<\frac{1}{\sqrt{M}}, \]
from which
\[ (x-2)^2<\frac{1}{M} \]
and hence
\[ \frac{1}{(x-2)^2}>M. \]
The definition is verified for every \(M>0\), so the limit is \(+\infty\).
Exercise 2 — level ★☆☆☆☆
Evaluate
\[ \lim_{x\to-3}\left(-\frac{4}{(x+3)^2}\right). \]
Answer
\[ \lim_{x\to-3}\left(-\frac{4}{(x+3)^2}\right)=-\infty. \]
Solution
As \(x\to-3\), we have
\[ x+3\to0. \]
The square \((x+3)^2\) is positive for every \(x\neq-3\) and tends to zero. More precisely,
\[ (x+3)^2\to0^+. \]
Consequently,
\[ \frac{4}{(x+3)^2}\to+\infty. \]
In the expression given in the exercise, however, a minus sign is present. Multiplying an arbitrarily large positive quantity by \(-1\) produces a quantity that is negative and arbitrarily large in absolute value. Hence
\[ -\frac{4}{(x+3)^2}\to-\infty. \]
We conclude that
\[ \lim_{x\to-3}\left(-\frac{4}{(x+3)^2}\right)=-\infty. \]
Since the same behaviour occurs both as \(x\to-3^-\) and as \(x\to-3^+\), the limit as \(x\to-3\) exists as an infinite limit and equals \(-\infty\). The line \(x=-3\) is furthermore a vertical asymptote.
Exercise 3 — level ★☆☆☆☆
Evaluate the left-hand and right-hand limits
\[ \lim_{x\to1^-}\frac{1}{x-1}, \qquad \lim_{x\to1^+}\frac{1}{x-1}, \]
and determine whether the limit as \(x\to1\) exists.
Answer
\[ \lim_{x\to1^-}\frac{1}{x-1}=-\infty, \qquad \lim_{x\to1^+}\frac{1}{x-1}=+\infty. \]
The limit as \(x\to1\) does not exist. The line \(x=1\) is a vertical asymptote.
Solution
The denominator \(x-1\) tends to zero as \(x\to1\). To determine the behaviour of the reciprocal, it is not enough to know that the denominator tends to zero: we must also establish whether it tends to \(0^-\) or to \(0^+\).
Consider first \(x\to1^-\). In this case \(x<1\), so
\[ x-1<0. \]
Moreover \(x-1\) approaches zero. We may thus write
\[ x-1\to0^-. \]
The reciprocal of a negative quantity tending to zero takes negative values that are arbitrarily large in absolute value. Hence
\[ \lim_{x\to1^-}\frac{1}{x-1}=-\infty. \]
Now consider \(x\to1^+\). This time \(x>1\), so
\[ x-1>0 \]
and hence
\[ x-1\to0^+. \]
It follows that
\[ \lim_{x\to1^+}\frac{1}{x-1}=+\infty. \]
The left-hand and right-hand limits have opposite signs. For the limit as \(x\to1\) to equal \(+\infty\), both the left-hand and right-hand limits would need to equal \(+\infty\); likewise, to obtain \(-\infty\), both would need to equal \(-\infty\).
Since this does not occur, the limit
\[ \lim_{x\to1}\frac{1}{x-1} \]
does not exist. Nevertheless, both the left-hand and right-hand limits are infinite, so \(x=1\) is a vertical asymptote.
Exercise 4 — level ★★☆☆☆
Study the behaviour of the function
\[ f(x)=\frac{2x+1}{x-2} \]
as \(x\to2^-\) and as \(x\to2^+\), and hence determine whether \(x=2\) is a vertical asymptote.
Answer
\[ \lim_{x\to2^-}\frac{2x+1}{x-2}=-\infty, \qquad \lim_{x\to2^+}\frac{2x+1}{x-2}=+\infty. \]
The limit as \(x\to2\) does not exist, and \(x=2\) is a vertical asymptote.
Solution
In this limit the numerator and denominator behave differently. As \(x\to2\),
\[ 2x+1\to5. \]
The numerator therefore tends to a positive number and, in particular, remains positive for all \(x\) sufficiently close to \(2\).
The denominator, on the other hand, tends to zero:
\[ x-2\to0. \]
Its sign depends on the side from which \(x\) approaches \(2\).
If \(x\to2^-\), then \(x<2\), so
\[ x-2\to0^-. \]
We thus have a positive numerator tending to \(5\) and a negative denominator tending to zero. The quotient is therefore negative and grows without bound in absolute value:
\[ \lim_{x\to2^-}\frac{2x+1}{x-2}=-\infty. \]
If instead \(x\to2^+\), then
\[ x-2\to0^+. \]
The numerator is still positive, while the denominator is positive and arbitrarily small. It follows that
\[ \lim_{x\to2^+}\frac{2x+1}{x-2}=+\infty. \]
Since the left-hand and right-hand limits are both infinite but of opposite sign, the limit as \(x\to2\) does not exist. The line \(x=2\) is nonetheless a vertical asymptote, since it suffices that at least one of the two one-sided limits be infinite.
Exercise 5 — level ★★☆☆☆
Evaluate
\[ \lim_{x\to1^+}\frac{1}{\sqrt{x-1}} \]
and give a geometric interpretation of the result.
Answer
\[ \lim_{x\to1^+}\frac{1}{\sqrt{x-1}}=+\infty. \]
The line \(x=1\) is a vertical asymptote.
Solution
The function
\[ f(x)=\frac{1}{\sqrt{x-1}} \]
is defined only when \(x-1>0\), that is, for \(x>1\). There are therefore no points of the domain arbitrarily close to \(1\) from the left: the limit can only be studied as \(x\to1^+\).
As \(x\to1^+\), we have
\[ x-1\to0^+. \]
The square root preserves positivity, so
\[ \sqrt{x-1}\to0^+. \]
The reciprocal of a positive quantity tending to zero tends to \(+\infty\). Hence
\[ \lim_{x\to1^+}\frac{1}{\sqrt{x-1}}=+\infty. \]
It is worth noting that a vertical asymptote does not require infinite behaviour on both sides. It is enough for either the right-hand limit or the left-hand limit to be infinite.
Consequently, the line
\[ x=1 \]
is a vertical asymptote of the graph of the function.
Exercise 6 — level ★★☆☆☆
Evaluate
\[ \lim_{x\to+\infty}\left(3+\frac{2}{x}\right) \]
and verify the result using the definition of a finite limit as \(x\to+\infty\).
Answer
\[ \lim_{x\to+\infty}\left(3+\frac{2}{x}\right)=3. \]
The line \(y=3\) is a horizontal asymptote as \(x\to+\infty\).
Solution
As \(x\to+\infty\), the denominator \(x\) grows without bound. Consequently,
\[ \frac{2}{x}\to0. \]
Adding \(3\), we immediately obtain
\[ 3+\frac{2}{x}\to3. \]
Let us now verify this result from the definition. We must show that, for every \(\varepsilon>0\), there exists \(A\in\mathbb{R}\) such that
\[ x>A \]
implies
\[ \left|3+\frac{2}{x}-3\right|<\varepsilon. \]
For \(x>0\),
\[ \left|3+\frac{2}{x}-3\right| = \left|\frac{2}{x}\right| = \frac{2}{x}. \]
We therefore require
\[ \frac{2}{x}<\varepsilon. \]
Since \(x>0\) and \(\varepsilon>0\), this inequality is equivalent to
\[ x>\frac{2}{\varepsilon}. \]
It therefore suffices to choose
\[ A=\frac{2}{\varepsilon}. \]
For every \(x>A\) we indeed have
\[ \left|3+\frac{2}{x}-3\right|<\varepsilon. \]
The definition is verified and the limit is \(3\). Geometrically, this means that the graph approaches the line \(y=3\) as \(x\) takes arbitrarily large positive values.
Exercise 7 — level ★★☆☆☆
Evaluate both limits
\[ \lim_{x\to+\infty}\frac{x}{\sqrt{x^2+1}}, \qquad \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}, \]
and determine any horizontal asymptotes.
Answer
\[ \lim_{x\to+\infty}\frac{x}{\sqrt{x^2+1}}=1, \qquad \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1. \]
The line \(y=1\) is the horizontal asymptote as \(x\to+\infty\), while the line \(y=-1\) is the horizontal asymptote as \(x\to-\infty\).
Solution
The key point to remember is that
\[ \sqrt{x^2}=|x|, \]
and not simply \(x\).
We rewrite the denominator by factoring out \(x^2\) from under the root:
\[ \sqrt{x^2+1} = \sqrt{x^2\left(1+\frac{1}{x^2}\right)} = |x|\sqrt{1+\frac{1}{x^2}}. \]
For \(x\to+\infty\), \(x\) is eventually positive, so \(|x|=x\). We obtain
\[ \frac{x}{\sqrt{x^2+1}} = \frac{x}{x\sqrt{1+\displaystyle\frac{1}{x^2}}} = \frac{1}{\sqrt{1+\displaystyle\frac{1}{x^2}}}. \]
Since
\[ \frac{1}{x^2}\to0, \]
it follows that
\[ \lim_{x\to+\infty}\frac{x}{\sqrt{x^2+1}}=1. \]
For \(x\to-\infty\), on the other hand, \(x\) is eventually negative, so \(|x|=-x\). Hence
\[ \frac{x}{\sqrt{x^2+1}} = \frac{x}{-x\sqrt{1+\displaystyle\frac{1}{x^2}}} = -\frac{1}{\sqrt{1+\displaystyle\frac{1}{x^2}}}. \]
Passing to the limit,
\[ \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+1}}=-1. \]
The two values are different. The graph therefore has two distinct horizontal asymptotes: \(y=1\) as \(x\to+\infty\) and \(y=-1\) as \(x\to-\infty\).
Exercise 8 — level ★★☆☆☆
Evaluate
\[ \lim_{x\to+\infty}\left(-2x^5+3x^2-1\right) \]
and
\[ \lim_{x\to-\infty}\left(-2x^5+3x^2-1\right). \]
Answer
\[ \lim_{x\to+\infty}\left(-2x^5+3x^2-1\right)=-\infty, \]
\[ \lim_{x\to-\infty}\left(-2x^5+3x^2-1\right)=+\infty. \]
Solution
For a polynomial, as \(x\to\pm\infty\), the overall behaviour is determined by the term of highest degree. Here the dominant term is
\[ -2x^5. \]
We can see this behaviour formally by factoring out \(x^5\):
\[ -2x^5+3x^2-1 = x^5\left(-2+\frac{3}{x^3}-\frac{1}{x^5}\right). \]
As \(x\to\pm\infty\),
\[ \frac{3}{x^3}\to0, \qquad \frac{1}{x^5}\to0, \]
so the factor in brackets tends to \(-2\), which is negative.
For \(x\to+\infty\),
\[ x^5\to+\infty. \]
Multiplying an unboundedly growing positive quantity by a factor tending to the negative number \(-2\), we obtain
\[ -2x^5+3x^2-1\to-\infty. \]
For \(x\to-\infty\), since the exponent \(5\) is odd,
\[ x^5\to-\infty. \]
The product of a negative quantity, arbitrarily large in absolute value, and a factor that is eventually negative is positive and grows without bound. Hence
\[ -2x^5+3x^2-1\to+\infty. \]
Exercise 9 — level ★★★☆☆
Evaluate
\[ \lim_{x\to\pm\infty}\frac{4x+1}{x^2+3}. \]
Answer
\[ \lim_{x\to+\infty}\frac{4x+1}{x^2+3}=0, \qquad \lim_{x\to-\infty}\frac{4x+1}{x^2+3}=0. \]
The line \(y=0\) is a horizontal asymptote in both directions.
Solution
Both numerator and denominator grow in absolute value, so a purely symbolic reading would suggest an indeterminate form of type
\[ \frac{\infty}{\infty}. \]
This form does not determine the value of the limit. We must compare the dominant terms.
The numerator has degree \(1\), while the denominator has degree \(2\). The denominator therefore grows faster.
To make the calculation explicit, we divide numerator and denominator by \(x^2\), the highest power present in the denominator:
\[ \frac{4x+1}{x^2+3} = \frac{\displaystyle\frac{4}{x}+\displaystyle\frac{1}{x^2}} {1+\displaystyle\frac{3}{x^2}}. \]
As \(x\to+\infty\) and as \(x\to-\infty\),
\[ \frac{1}{x}\to0, \qquad \frac{1}{x^2}\to0. \]
The numerator therefore tends to \(0\), while the denominator tends to \(1\). It follows that
\[ \frac{\displaystyle\frac{4}{x}+\displaystyle\frac{1}{x^2}} {1+\displaystyle\frac{3}{x^2}} \to0. \]
Hence both limits equal \(0\), and the line \(y=0\) is a horizontal asymptote both as \(x\to+\infty\) and as \(x\to-\infty\).
Exercise 10 — level ★★★☆☆
Evaluate
\[ \lim_{x\to\pm\infty}\frac{3x^2-x+1}{2x^2+5}. \]
Answer
\[ \lim_{x\to+\infty}\frac{3x^2-x+1}{2x^2+5} = \lim_{x\to-\infty}\frac{3x^2-x+1}{2x^2+5} = \frac{3}{2}. \]
Solution
Both the numerator and the denominator are polynomials of degree \(2\). For a rational function whose numerator and denominator have the same degree, the behaviour at infinity is determined by the ratio of the coefficients of the highest-degree terms.
Let us verify this result algebraically. We divide every term by \(x^2\):
\[ \frac{3x^2-x+1}{2x^2+5} = \frac{3-\displaystyle\frac{1}{x}+\displaystyle\frac{1}{x^2}} {2+\displaystyle\frac{5}{x^2}}. \]
As \(x\to+\infty\) or \(x\to-\infty\),
\[ \frac{1}{x}\to0, \qquad \frac{1}{x^2}\to0. \]
Consequently, the numerator tends to \(3\) and the denominator tends to \(2\):
\[ 3-\frac{1}{x}+\frac{1}{x^2}\to3, \qquad 2+\frac{5}{x^2}\to2. \]
Since the limit of the denominator is nonzero, we may pass to the quotient:
\[ \lim_{x\to\pm\infty}\frac{3x^2-x+1}{2x^2+5} = \frac{3}{2}. \]
The line
\[ y=\frac{3}{2} \]
is therefore a horizontal asymptote in both directions.
Exercise 11 — level ★★★☆☆
Evaluate
\[ \lim_{x\to+\infty}\frac{2x^3+1}{x^2+1} \]
and
\[ \lim_{x\to-\infty}\frac{2x^3+1}{x^2+1}. \]
Answer
\[ \lim_{x\to+\infty}\frac{2x^3+1}{x^2+1}=+\infty, \qquad \lim_{x\to-\infty}\frac{2x^3+1}{x^2+1}=-\infty. \]
Solution
The numerator has degree \(3\), while the denominator has degree \(2\). Thus the degree of the numerator exceeds that of the denominator by one.
We divide numerator and denominator by \(x^2\):
\[ \frac{2x^3+1}{x^2+1} = \frac{2x+\displaystyle\frac{1}{x^2}} {1+\displaystyle\frac{1}{x^2}}. \]
As \(x\to+\infty\),
\[ 2x+\frac{1}{x^2}\to+\infty, \qquad 1+\frac{1}{x^2}\to1. \]
The denominator therefore tends to a positive nonzero number, while the numerator tends to \(+\infty\). Hence
\[ \lim_{x\to+\infty}\frac{2x^3+1}{x^2+1}=+\infty. \]
For \(x\to-\infty\), on the other hand,
\[ 2x+\frac{1}{x^2}\to-\infty, \]
while the denominator still tends to \(1>0\). Consequently,
\[ \lim_{x\to-\infty}\frac{2x^3+1}{x^2+1}=-\infty. \]
This result is consistent with comparing the dominant terms:
\[ \frac{2x^3}{x^2}=2x. \]
For large values of \(|x|\), the ratio thus behaves essentially like \(2x\).
Exercise 12 — level ★★★☆☆
Evaluate
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+1}-x\right). \]
Answer
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+1}-x\right)=0. \]
Solution
If we consider the two terms separately, as \(x\to+\infty\) we have
\[ \sqrt{x^2+1}\to+\infty, \qquad x\to+\infty. \]
The indeterminate form
\[ \infty-\infty \]
therefore arises. We cannot draw a conclusion directly. Whenever a difference involving a square root appears, it is natural to rationalise by multiplying and dividing by the conjugate expression:
\[ \sqrt{x^2+1}-x = \frac{\left(\sqrt{x^2+1}-x\right)\left(\sqrt{x^2+1}+x\right)} {\sqrt{x^2+1}+x}. \]
In the numerator we use the difference of two squares:
\[ \left(\sqrt{x^2+1}\right)^2-x^2 = x^2+1-x^2 = 1. \]
Hence
\[ \sqrt{x^2+1}-x = \frac{1}{\sqrt{x^2+1}+x}. \]
As \(x\to+\infty\), both terms in the denominator are positive and grow without bound, so
\[ \sqrt{x^2+1}+x\to+\infty. \]
The reciprocal of a quantity tending to \(+\infty\) tends to zero. It follows that
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+1}-x\right)=0. \]
Rationalisation has thus turned the indeterminate form \(\infty-\infty\) into a quotient whose behaviour can be determined at once.
Exercise 13 — level ★★★★☆
Evaluate
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+3x}-x\right). \]
Answer
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+3x}-x\right)=\frac{3}{2}. \]
Solution
As \(x\to+\infty\), both \(\sqrt{x^2+3x}\) and \(x\) tend to \(+\infty\). We are again faced with the indeterminate form
\[ \infty-\infty. \]
We rationalise:
\[ \sqrt{x^2+3x}-x = \frac{\left(\sqrt{x^2+3x}-x\right) \left(\sqrt{x^2+3x}+x\right)} {\sqrt{x^2+3x}+x}. \]
The numerator becomes
\[ x^2+3x-x^2=3x. \]
Hence
\[ \sqrt{x^2+3x}-x = \frac{3x}{\sqrt{x^2+3x}+x}. \]
We now factor \(x^2\) out from under the root:
\[ \sqrt{x^2+3x} = \sqrt{x^2\left(1+\frac{3}{x}\right)} = |x|\sqrt{1+\frac{3}{x}}. \]
Since we are considering \(x\to+\infty\), \(x\) is eventually positive, so \(|x|=x\). We obtain
\[ \sqrt{x^2+3x} = x\sqrt{1+\frac{3}{x}}. \]
Substituting,
\[ \frac{3x}{\sqrt{x^2+3x}+x} = \frac{3x} {x\sqrt{1+\displaystyle\frac{3}{x}}+x}. \]
We factor \(x\) out of the denominator and simplify:
\[ \frac{3} {\sqrt{1+\displaystyle\frac{3}{x}}+1}. \]
Since
\[ \frac{3}{x}\to0, \]
we obtain
\[ \lim_{x\to+\infty} \frac{3} {\sqrt{1+\displaystyle\frac{3}{x}}+1} = \frac{3}{1+1} = \frac{3}{2}. \]
Exercise 14 — level ★★★★☆
Evaluate
\[ \lim_{x\to-\infty}\left(\sqrt{x^2+1}+x\right). \]
Answer
\[ \lim_{x\to-\infty}\left(\sqrt{x^2+1}+x\right)=0. \]
Solution
As \(x\to-\infty\), we have
\[ \sqrt{x^2+1}\to+\infty, \qquad x\to-\infty. \]
The sum therefore presents the indeterminate form
\[ \infty-\infty. \]
We rationalise by multiplying and dividing by the conjugate expression \(\sqrt{x^2+1}-x\):
\[ \sqrt{x^2+1}+x = \frac{\left(\sqrt{x^2+1}+x\right) \left(\sqrt{x^2+1}-x\right)} {\sqrt{x^2+1}-x}. \]
In the numerator we obtain
\[ x^2+1-x^2=1. \]
Hence
\[ \sqrt{x^2+1}+x = \frac{1}{\sqrt{x^2+1}-x}. \]
Let us examine the denominator. As \(x\to-\infty\),
\[ \sqrt{x^2+1}\to+\infty \]
and, since \(x\to-\infty\),
\[ -x\to+\infty. \]
Hence
\[ \sqrt{x^2+1}-x = \sqrt{x^2+1}+(-x)\to+\infty. \]
Its reciprocal then tends to zero:
\[ \frac{1}{\sqrt{x^2+1}-x}\to0. \]
We conclude that
\[ \lim_{x\to-\infty}\left(\sqrt{x^2+1}+x\right)=0. \]
Exercise 15 — level ★★★★☆
Using the comparison theorem, evaluate
\[ \lim_{x\to+\infty}\left(x+2\sin(x)\right). \]
Answer
\[ \lim_{x\to+\infty}\left(x+2\sin(x)\right)=+\infty. \]
Solution
The presence of the term \(\sin(x)\) means that the function oscillates. This, however, does not prevent an infinite limit from existing.
For every \(x\in\mathbb{R}\),
\[ -1\leq\sin(x)\leq1. \]
Multiplying by \(2\),
\[ -2\leq2\sin(x)\leq2. \]
Adding \(x\) to every term,
\[ x-2\leq x+2\sin(x)\leq x+2. \]
To show that the function tends to \(+\infty\), it is enough to use the lower bound:
\[ x+2\sin(x)\geq x-2. \]
Since
\[ \lim_{x\to+\infty}(x-2)=+\infty, \]
the comparison theorem gives
\[ \lim_{x\to+\infty}\left(x+2\sin(x)\right)=+\infty. \]
It is important to understand what has happened here. The term \(2\sin(x)\) continues to oscillate between \(-2\) and \(2\), but these oscillations are bounded. The term \(x\), by contrast, grows without bound. Even in the least favourable case, that is, when \(2\sin(x)=-2\), the function is still at least equal to \(x-2\), which nonetheless tends to \(+\infty\).
Exercise 16 — level ★★★★☆
Using the squeeze theorem, evaluate
\[ \lim_{x\to+\infty}\left(-x+3\cos(x)\right). \]
Answer
\[ \lim_{x\to+\infty}\left(-x+3\cos(x)\right)=-\infty. \]
Solution
Here too an oscillating term appears. We know that
\[ -1\leq\cos(x)\leq1. \]
Multiplying by \(3\),
\[ -3\leq3\cos(x)\leq3. \]
Adding \(-x\),
\[ -x-3\leq -x+3\cos(x)\leq -x+3. \]
To show that the function tends to \(-\infty\), we need an upper bound that itself tends to \(-\infty\). From the inequality above,
\[ -x+3\cos(x)\leq -x+3. \]
Moreover,
\[ \lim_{x\to+\infty}(-x+3)=-\infty. \]
By the comparison theorem,
\[ \lim_{x\to+\infty}\left(-x+3\cos(x)\right)=-\infty. \]
The oscillating term always remains between \(-3\) and \(3\), while the term \(-x\) decreases without bound. The oscillations are therefore not enough to alter the dominant behaviour of the function.
Exercise 17 — level ★★★★☆
Evaluate
\[ \lim_{x\to0}\frac{3+\cos\left(\displaystyle\frac{1}{x}\right)}{x^2}. \]
Answer
\[ \lim_{x\to0}\frac{3+\cos\left(\displaystyle\frac{1}{x}\right)}{x^2}=+\infty. \]
Solution
The term
\[ \cos\left(\frac{1}{x}\right) \]
oscillates ever more rapidly as \(x\to0\) and has no limit. On its own, however, this does not mean that the whole function fails to have a limit.
We use the fact that, for every real value of the argument,
\[ -1\leq\cos\left(\frac{1}{x}\right)\leq1. \]
Adding \(3\),
\[ 2\leq3+\cos\left(\frac{1}{x}\right)\leq4. \]
For \(x\neq0\), we have \(x^2>0\). Dividing by \(x^2\), the direction of the inequalities is unchanged:
\[ \frac{2}{x^2} \leq \frac{3+\cos\left(\displaystyle\frac{1}{x}\right)}{x^2} \leq \frac{4}{x^2}. \]
To show that the limit equals \(+\infty\), it suffices to use the lower bound:
\[ \frac{3+\cos\left(\displaystyle\frac{1}{x}\right)}{x^2} \geq \frac{2}{x^2}. \]
Since
\[ \lim_{x\to0}\frac{2}{x^2}=+\infty, \]
the squeeze theorem gives
\[ \lim_{x\to0}\frac{3+\cos\left(\displaystyle\frac{1}{x}\right)}{x^2}=+\infty. \]
This exercise shows once again that the existence of an infinite limit does not require monotonicity and is not prevented by the presence of oscillations, provided that the function can be guaranteed to exceed every positive threshold eventually.
Exercise 18 — level ★★★★☆
Study the limit
\[ \lim_{x\to1}\frac{x^2-1}{x-1} \]
and determine whether \(x=1\) is a vertical asymptote.
Answer
\[ \lim_{x\to1}\frac{x^2-1}{x-1}=2. \]
The line \(x=1\) is not a vertical asymptote.
Solution
Substituting \(x=1\) formally, both numerator and denominator vanish:
\[ x^2-1\to0, \qquad x-1\to0. \]
The indeterminate form
\[ \frac{0}{0} \]
therefore arises. We cannot conclude that the limit is infinite simply because the denominator tends to zero. We must first analyse the expression.
We factorise the numerator using the difference of two squares:
\[ x^2-1=(x-1)(x+1). \]
For \(x\neq1\),
\[ \frac{x^2-1}{x-1} = \frac{(x-1)(x+1)}{x-1} = x+1. \]
In evaluating the limit we may use this identity, because the limit depends on the behaviour of the function at points arbitrarily close to \(1\), not on the value taken at the point \(x=1\) itself.
Hence
\[ \lim_{x\to1}\frac{x^2-1}{x-1} = \lim_{x\to1}(x+1) = 2. \]
The limit is finite. Consequently \(x=1\) is not a vertical asymptote.
This exercise highlights an important principle: the vanishing of the denominator alone is not sufficient to conclude that a vertical asymptote exists. The limit must always be evaluated explicitly.
Exercise 19 — level ★★★★★
Study the limit
\[ \lim_{x\to1}\frac{x^2+1}{(x-1)^2} \]
also evaluating the left-hand and right-hand limits and determining any vertical asymptote.
Answer
\[ \lim_{x\to1^-}\frac{x^2+1}{(x-1)^2}=+\infty, \qquad \lim_{x\to1^+}\frac{x^2+1}{(x-1)^2}=+\infty. \]
Consequently,
\[ \lim_{x\to1}\frac{x^2+1}{(x-1)^2}=+\infty, \]
and \(x=1\) is a vertical asymptote.
Solution
Let us analyse the numerator and denominator separately. As \(x\to1\),
\[ x^2+1\to2. \]
The numerator therefore tends to a strictly positive number and remains positive for all \(x\) sufficiently close to \(1\).
The denominator satisfies
\[ (x-1)^2\to0. \]
Being a square,
\[ (x-1)^2>0 \]
for every \(x\neq1\). The denominator therefore always tends to zero through positive values, regardless of the side of approach:
\[ (x-1)^2\to0^+ \]
both as \(x\to1^-\) and as \(x\to1^+\).
For the left-hand limit we thus have a numerator tending to \(2>0\) and a denominator tending to \(0^+\). It follows that
\[ \lim_{x\to1^-}\frac{x^2+1}{(x-1)^2}=+\infty. \]
The same reasoning holds from the right:
\[ \lim_{x\to1^+}\frac{x^2+1}{(x-1)^2}=+\infty. \]
Since both the left-hand and right-hand limits equal \(+\infty\), we may conclude that
\[ \lim_{x\to1}\frac{x^2+1}{(x-1)^2}=+\infty. \]
The line
\[ x=1 \]
is therefore a vertical asymptote.
Exercise 20 — level ★★★★★
Consider the function
\[ f(x)=\frac{2x^3+1}{x^2-4}. \]
Evaluate the limits as \(x\to2^-\), \(x\to2^+\), \(x\to-2^-\), \(x\to-2^+\), \(x\to+\infty\) and \(x\to-\infty\). Then determine any vertical and horizontal asymptotes.
Answer
\[ \lim_{x\to2^-}f(x)=-\infty, \qquad \lim_{x\to2^+}f(x)=+\infty, \]
\[ \lim_{x\to-2^-}f(x)=-\infty, \qquad \lim_{x\to-2^+}f(x)=+\infty, \]
\[ \lim_{x\to+\infty}f(x)=+\infty, \qquad \lim_{x\to-\infty}f(x)=-\infty. \]
The lines \(x=-2\) and \(x=2\) are vertical asymptotes. The function has no horizontal asymptotes.
Solution
The denominator can be factorised:
\[ x^2-4=(x-2)(x+2). \]
The points \(x=2\) and \(x=-2\) are therefore the only points at which the denominator vanishes. To establish the behaviour of the function near these points we must examine the signs carefully.
Let us start with \(x=2\). The numerator tends to
\[ 2\cdot2^3+1=17, \]
so it is eventually positive near \(2\).
As \(x\to2^-\), the factor \(x-2\) is negative and tends to zero, while \(x+2\) is positive and tends to \(4\). Consequently,
\[ (x-2)(x+2)\to0^-. \]
A positive numerator divided by a negative quantity tending to zero gives
\[ \lim_{x\to2^-}\frac{2x^3+1}{x^2-4}=-\infty. \]
As \(x\to2^+\), both factors \(x-2\) and \(x+2\) are positive, so
\[ x^2-4\to0^+. \]
It follows that
\[ \lim_{x\to2^+}\frac{2x^3+1}{x^2-4}=+\infty. \]
Let us now turn to \(x=-2\). The numerator tends to
\[ 2(-2)^3+1=-16+1=-15, \]
so it is eventually negative near \(-2\).
As \(x\to-2^-\), we have \(x+2<0\) and \(x-2<0\). The product of the two factors is positive:
\[ (x-2)(x+2)\to0^+. \]
Dividing a negative numerator by a positive quantity tending to zero gives
\[ \lim_{x\to-2^-}\frac{2x^3+1}{x^2-4}=-\infty. \]
As \(x\to-2^+\), on the other hand, \(x+2>0\) while \(x-2<0\). The denominator therefore tends to zero through negative values:
\[ (x-2)(x+2)\to0^-. \]
The quotient of a negative numerator by a negative denominator is positive, so
\[ \lim_{x\to-2^+}\frac{2x^3+1}{x^2-4}=+\infty. \]
We have thus already established that both lines
\[ x=-2 \qquad\text{and}\qquad x=2 \]
are vertical asymptotes.
Let us now examine the behaviour at infinity. The numerator has degree \(3\), while the denominator has degree \(2\). We divide numerator and denominator by \(x^2\):
\[ \frac{2x^3+1}{x^2-4} = \frac{2x+\displaystyle\frac{1}{x^2}} {1-\displaystyle\frac{4}{x^2}}. \]
As \(x\to+\infty\),
\[ 2x+\frac{1}{x^2}\to+\infty \]
and
\[ 1-\frac{4}{x^2}\to1. \]
The denominator tends to a positive nonzero number, so
\[ \lim_{x\to+\infty}\frac{2x^3+1}{x^2-4}=+\infty. \]
As \(x\to-\infty\),
\[ 2x+\frac{1}{x^2}\to-\infty, \]
while the denominator still tends to \(1\). Hence
\[ \lim_{x\to-\infty}\frac{2x^3+1}{x^2-4}=-\infty. \]
Since the function tends to a real number neither as \(x\to+\infty\) nor as \(x\to-\infty\), there are no horizontal asymptotes.
This exercise brings together the main ideas used in the study of infinite limits and limits at infinity: factorisation, checking the sign in one-sided limits, identifying vertical asymptotes, and comparing dominant terms in limits at infinity.