The following exercises are designed to help you recognise and resolve the principal indeterminate forms that arise in the computation of limits. For each limit we shall first identify the indeterminate form, then select an appropriate transformation and justify every step in detail.
The exercises are arranged in order of increasing difficulty and make use exclusively of the tools introduced in the theory: factorisation and simplification, rationalisation, division by the dominant term, standard limits, asymptotic equivalences, the squeeze theorem, and transformation by means of the logarithm.
Exercise 1 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to 2}\frac{x^2-4}{x-2}. \]
Answer
\[ \lim_{x\to 2}\frac{x^2-4}{x-2}=4. \]
Solution
Evaluating the limits of the numerator and denominator separately, we obtain
\[ \lim_{x\to2}(x^2-4)=0 \qquad\text{and}\qquad \lim_{x\to2}(x-2)=0. \]
The limit therefore presents the indeterminate form \(\displaystyle\frac{0}{0}\).
The numerator is a difference of squares:
\[ x^2-4=(x-2)(x+2). \]
For \(x\neq 2\), a condition that is sufficient for the purposes of the limit as \(x\to2\), we may cancel the common factor:
\[ \frac{x^2-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2. \]
The two expressions coincide at every point of the domain sufficiently close to \(2\), possibly excluding \(2\) itself, and therefore share the same limit. Hence
\[ \lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4. \]
Exercise 2 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}. \]
Answer
\[ \lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}=\frac14. \]
Solution
As \(x\to0\), the numerator tends to \(\sqrt4-2=0\) and the denominator tends to zero. We thus obtain the form \(\displaystyle\frac00\). The presence of a difference involving a square root suggests multiplying by the conjugate expression.
For \(x\neq0\), we multiply numerator and denominator by \(\sqrt{4+x}+2\):
\[ \frac{\sqrt{4+x}-2}{x}\cdot\frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}=\frac{(4+x)-4}{x(\sqrt{4+x}+2)}. \]
In the numerator we are left with \(x\), which cancels with the denominator:
\[ \frac{x}{x(\sqrt{4+x}+2)}=\frac{1}{\sqrt{4+x}+2}. \]
The new expression exhibits no indeterminate form. By continuity of the square root and of the algebraic operations,
\[ \lim_{x\to0}\frac{1}{\sqrt{4+x}+2}=\frac{1}{2+2}=\frac14. \]
Exercise 3 — level ★☆☆☆☆
Evaluate the limit
\[ \lim_{x\to0}\frac{\sin(5x)}{x}. \]
Answer
\[ \lim_{x\to0}\frac{\sin(5x)}{x}=5. \]
Solution
Since \(\sin(5x)\to0\) and \(x\to0\), the limit has the form \(\displaystyle\frac00\). The standard limit \(\displaystyle\lim_{t\to0}\frac{\sin t}{t}=1\) cannot yet be applied directly, since the denominator contains \(x\) rather than \(5x\).
We multiply and divide by \(5\):
\[ \frac{\sin(5x)}{x}=5\frac{\sin(5x)}{5x}. \]
Setting, formally, \(t=5x\), we see that \(x\to0\) implies \(t\to0\). Hence
\[ \lim_{x\to0}5\frac{\sin(5x)}{5x}=5\cdot1=5. \]
Exercise 4 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to0}\frac{e^{3x}-1}{x}. \]
Answer
\[ \lim_{x\to0}\frac{e^{3x}-1}{x}=3. \]
Solution
The numerator tends to \(e^0-1=0\), while the denominator tends to zero: we are in the form \(\displaystyle\frac00\). We use the standard limit \(\displaystyle\lim_{t\to0}\frac{e^t-1}{t}=1\).
We write
\[ \frac{e^{3x}-1}{x}=3\frac{e^{3x}-1}{3x}. \]
Since \(3x\to0\), the quotient on the right tends to \(1\). It follows that
\[ \lim_{x\to0}\frac{e^{3x}-1}{x}=3\cdot1=3. \]
Exercise 5 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to0}\frac{\ln(1+4x)}{2x}. \]
Answer
\[ \lim_{x\to0}\frac{\ln(1+4x)}{2x}=2. \]
Solution
As \(x\to0\), we have \(\ln(1+4x)\to\ln1=0\) and \(2x\to0\). The limit presents the form \(\displaystyle\frac00\).
To reduce the expression to the standard limit \(\displaystyle\lim_{t\to0}\frac{\ln(1+t)}t=1\), we introduce the term \(4x\) in the denominator:
\[ \frac{\ln(1+4x)}{2x}=2\frac{\ln(1+4x)}{4x}. \]
Since \(4x\to0\), the second factor tends to \(1\). Therefore
\[ \lim_{x\to0}\frac{\ln(1+4x)}{2x}=2. \]
Exercise 6 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to0}\frac{x}{|x|}. \]
Answer
The limit as \(x\to0\) does not exist.
Solution
Both numerator and denominator tend to zero, so we obtain the form \(\displaystyle\frac00\). This, however, does not imply that the limit exists.
For \(x>0\), we have \(|x|=x\), so that
\[ \frac{x}{|x|}=1 \qquad\Longrightarrow\qquad \lim_{x\to0^+}\frac{x}{|x|}=1. \]
For \(x<0\), on the other hand, \(|x|=-x\), and thus
\[ \frac{x}{|x|}=\frac{x}{-x}=-1 \qquad\Longrightarrow\qquad \lim_{x\to0^-}\frac{x}{|x|}=-1. \]
Since the right-hand and left-hand limits are finite but distinct, the limit as \(x\to0\) does not exist.
Exercise 7 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\frac{4x^2-3x+1}{2x^2+x-5}. \]
Answer
\[ \lim_{x\to+\infty}\frac{4x^2-3x+1}{2x^2+x-5}=2. \]
Solution
Both numerator and denominator tend to \(+\infty\); the limit therefore has the form \(\displaystyle\frac{\infty}{\infty}\). To compare the dominant terms we divide numerator and denominator by \(x^2\), the highest power present:
\[ \frac{4x^2-3x+1}{2x^2+x-5}=\frac{4-\displaystyle\frac3x+\displaystyle\frac1{x^2}}{2+\displaystyle\frac1x-\displaystyle\frac5{x^2}}. \]
As \(x\to+\infty\), both \(\displaystyle\frac1x\) and \(\displaystyle\frac1{x^2}\) tend to zero. The denominator of the new expression tends to \(2\neq0\), so we may apply the algebra of limits:
\[ \lim_{x\to+\infty}\frac{4-\displaystyle\frac3x+\displaystyle\frac1{x^2}}{2+\displaystyle\frac1x-\displaystyle\frac5{x^2}}=\frac42=2. \]
Exercise 8 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\frac{3x+1}{x^2+4}. \]
Answer
\[ \lim_{x\to+\infty}\frac{3x+1}{x^2+4}=0. \]
Solution
Both numerator and denominator tend to \(+\infty\), so the form is \(\displaystyle\frac{\infty}{\infty}\). The denominator has higher degree than the numerator: we expect it to grow more rapidly, but this must be established rigorously.
We divide throughout by \(x^2\):
\[ \frac{3x+1}{x^2+4}=\frac{\displaystyle\frac3x+\displaystyle\frac1{x^2}}{1+\displaystyle\frac4{x^2}}. \]
The numerator tends to zero and the denominator tends to \(1\). Consequently,
\[ \lim_{x\to+\infty}\frac{3x+1}{x^2+4}=\frac01=0. \]
Exercise 9 — level ★★☆☆☆
Evaluate the limit
\[ \lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x}. \]
Answer
\[ \lim_{x\to-\infty}\frac{\sqrt{x^2+1}}{x}=-1. \]
Solution
Both numerator and denominator diverge in absolute value, so we obtain a form of type \(\displaystyle\frac{\infty}{\infty}\). We factor \(x^2\) out of the radical:
\[ \sqrt{x^2+1}=\sqrt{x^2\left(1+\frac1{x^2}\right)}=|x|\sqrt{1+\frac1{x^2}}. \]
It is essential not to replace \(\sqrt{x^2}\) by \(x\): the identity valid for every real number is \(\sqrt{x^2}=|x|\). As \(x\to-\infty\), we eventually have \(x<0\), so \(|x|=-x\). Hence
\[ \frac{\sqrt{x^2+1}}x=\frac{-x\sqrt{1+\displaystyle\frac1{x^2}}}{x}=-\sqrt{1+\frac1{x^2}}. \]
Since \(\displaystyle\frac1{x^2}\to0\), we obtain
\[ \lim_{x\to-\infty}\frac{\sqrt{x^2+1}}x=-\sqrt1=-1. \]
Exercise 10 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to0^+}x\ln x. \]
Answer
\[ \lim_{x\to0^+}x\ln x=0. \]
Solution
As \(x\to0^+\), we have \(x\to0\) and \(\ln x\to-\infty\). The product therefore has the form \(0\cdot(-\infty)\), which is an instance of the indeterminate form \(0\cdot\infty\).
We rewrite the product as a quotient:
\[ x\ln x=\frac{\ln x}{\displaystyle\frac1x}. \]
To make the behaviour more transparent, we set \(t=\displaystyle\frac1x\). If \(x\to0^+\), then \(t\to+\infty\), while \(x=\displaystyle\frac1t\) and \(\ln x=-\ln t\). Hence
\[ x\ln x=-\frac{\ln t}{t}. \]
By the fundamental hierarchy of infinities, the logarithm grows more slowly than any positive power; in particular, \(\ln t=o(t)\). It follows that
\[ \lim_{t\to+\infty}\frac{\ln t}{t}=0, \]
and therefore
\[ \lim_{x\to0^+}x\ln x=0. \]
The product takes negative values near zero, so it approaches zero from below; the value of the limit remains, in any case, \(0\).
Exercise 11 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\frac{\sin x}{x}\cdot\sqrt{x}. \]
Answer
\[ \lim_{x\to+\infty}\frac{\sin x}{x}\cdot\sqrt{x}=0. \]
Solution
The factor \(\displaystyle\frac{\sin x}{x}\) tends to zero, since \(|\sin x|\leq1\), while \(\sqrt{x}\to+\infty\). We thus obtain the form \(0\cdot\infty\).
We simplify the powers of \(x\):
\[ \frac{\sin x}{x}\sqrt{x}=\frac{\sin x}{\sqrt{x}}. \]
Since \(|\sin x|\leq1\), for \(x>0\) we have
\[ \left|\frac{\sin x}{\sqrt{x}}\right|\leq\frac1{\sqrt{x}}. \]
The term on the right tends to zero as \(x\to+\infty\). Equivalently,
\[ -\frac1{\sqrt{x}}\leq\frac{\sin x}{\sqrt{x}}\leq\frac1{\sqrt{x}}. \]
Both bounds tend to zero; by the squeeze theorem, the expression in the middle also tends to zero.
Exercise 12 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+3x}-x\right). \]
Answer
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+3x}-x\right)=\frac32. \]
Solution
Both terms of the difference tend to \(+\infty\), so we are in the presence of the form \(\infty-\infty\). We cannot formally subtract two infinities. We rationalise by multiplying and dividing by the conjugate expression:
\[ \sqrt{x^2+3x}-x=\frac{(\sqrt{x^2+3x}-x)(\sqrt{x^2+3x}+x)}{\sqrt{x^2+3x}+x}. \]
The numerator is a difference of squares:
\[ \sqrt{x^2+3x}-x=\frac{x^2+3x-x^2}{\sqrt{x^2+3x}+x}=\frac{3x}{\sqrt{x^2+3x}+x}. \]
As \(x\to+\infty\), we eventually have \(x>0\). We may therefore write
\[ \sqrt{x^2+3x}=x\sqrt{1+\frac3x}. \]
Dividing numerator and denominator by \(x\), we obtain
\[ \frac{3x}{\sqrt{x^2+3x}+x}=\frac3{\sqrt{1+\displaystyle\frac3x}+1}. \]
Now \(\displaystyle\frac3x\to0\), so
\[ \lim_{x\to+\infty}\frac3{\sqrt{1+\displaystyle\frac3x}+1}=\frac3{1+1}=\frac32. \]
Exercise 13 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right). \]
Answer
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=1. \]
Solution
As \(x\to+\infty\), both radicals tend to \(+\infty\). The difference thus presents the indeterminate form \(\infty-\infty\).
We multiply and divide by the conjugate expression:
\[ \sqrt{x^2+x}-\sqrt{x^2-x}=\frac{(\sqrt{x^2+x}-\sqrt{x^2-x})(\sqrt{x^2+x}+\sqrt{x^2-x})}{\sqrt{x^2+x}+\sqrt{x^2-x}}. \]
In the numerator we obtain a difference of squares:
\[ (x^2+x)-(x^2-x)=2x. \]
Hence
\[ \sqrt{x^2+x}-\sqrt{x^2-x}=\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}. \]
Since \(x>0\) eventually, we may extract \(x^2\) from both radicals using \(\sqrt{x^2}=|x|=x\):
\[ \frac{2x}{x\sqrt{1+\displaystyle\frac1x}+x\sqrt{1-\displaystyle\frac1x}}=\frac2{\sqrt{1+\displaystyle\frac1x}+\sqrt{1-\displaystyle\frac1x}}. \]
Finally, \(\displaystyle\frac1x\to0\), so the denominator tends to \(1+1=2\) and
\[ \lim_{x\to+\infty}\left(\sqrt{x^2+x}-\sqrt{x^2-x}\right)=\frac22=1. \]
Exercise 14 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}\left(1+\frac2x\right)^{3x}. \]
Answer
\[ \lim_{x\to+\infty}\left(1+\frac2x\right)^{3x}=e^6. \]
Solution
The base tends to \(1\), while the exponent tends to \(+\infty\). The limit presents the form \(1^\infty\).
Set
\[ y(x)=\left(1+\frac2x\right)^{3x}. \]
The base is positive for \(x\) sufficiently large, so we may take the logarithm:
\[ \ln y(x)=3x\ln\left(1+\frac2x\right). \]
To make use of the standard limit for the logarithm, we write
\[ 3x\ln\left(1+\frac2x\right)=6\frac{\ln\left(1+\displaystyle\frac2x\right)}{\displaystyle\frac2x}. \]
Since \(\displaystyle\frac2x\to0\), the quotient on the right tends to \(1\); hence \(\ln y(x)\to6\). By continuity of the exponential function,
\[ y(x)=e^{\ln y(x)}\to e^6. \]
Exercise 15 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to0}(1+\sin x)^{\frac1x}. \]
Answer
\[ \lim_{x\to0}(1+\sin x)^{\frac1x}=e. \]
Solution
As \(x\to0\), the base tends to \(1\). The exponent tends to \(+\infty\) as \(x\to0^+\) and to \(-\infty\) as \(x\to0^-\). Thus the right-hand and left-hand limits exhibit the indeterminate forms \(1^{+\infty}\) and \(1^{-\infty}\), respectively. Both can be treated by means of the logarithm.
The base is positive in a neighbourhood of zero. Setting \(y(x)=(1+\sin x)^{1/x}\), we obtain
\[ \ln y(x)=\frac{\ln(1+\sin x)}x. \]
We split the quotient into two factors:
\[ \frac{\ln(1+\sin x)}x=\frac{\ln(1+\sin x)}{\sin x}\cdot\frac{\sin x}{x}. \]
Since \(\sin x\to0\), the first factor tends to \(1\) by the standard limit for the logarithm; the second likewise tends to \(1\) by the standard limit for the sine. Hence
\[ \lim_{x\to0}\ln y(x)=1. \]
Applying the exponential function, we obtain
\[ \lim_{x\to0^+}y(x)=e \qquad\text{and}\qquad \lim_{x\to0^-}y(x)=e. \]
Since the right-hand and left-hand limits coincide, we conclude that
\[ \lim_{x\to0}y(x)=e. \]
Exercise 16 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to0^+}x^x. \]
Answer
\[ \lim_{x\to0^+}x^x=1. \]
Solution
The base tends to \(0^+\) and the exponent tends to zero. We thus obtain the indeterminate form \(0^0\). Since \(x>0\), we may set \(y(x)=x^x\) and take the logarithm:
\[ \ln y(x)=x\ln x. \]
The product on the right has the form \(0\cdot(-\infty)\). As shown in Exercise 10,
\[ \lim_{x\to0^+}x\ln x=0. \]
Consequently, \(\ln y(x)\to0\). By continuity of the exponential function,
\[ \lim_{x\to0^+}x^x=e^0=1. \]
The form \(0^0\) has not been treated as an ordinary power: the result was obtained by studying the product of the exponent and the logarithm of the base.
Exercise 17 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to0^+}x^{\frac1{\ln x}}. \]
Answer
\[ \lim_{x\to0^+}x^{\frac1{\ln x}}=e. \]
Solution
As \(x\to0^+\), the base tends to zero and \(\ln x\to-\infty\), so the exponent \(\displaystyle\frac1{\ln x}\) tends to zero. The form is \(0^0\).
For \(0<x<1\), both the base and the logarithm are well defined. Set \(y(x)=x^{1/\ln x}\). Then
\[ \ln y(x)=\frac1{\ln x}\ln x=1. \]
In fact, this equality is exact for every \(x>0\) with \(x\neq1\). Hence
\[ y(x)=e^{\ln y(x)}=e, \]
and the function is constantly equal to \(e\) on the portion of the domain under consideration. It follows immediately that the limit equals \(e\).
Exercise 18 — level ★★★☆☆
Evaluate the limit
\[ \lim_{x\to+\infty}x^{\frac1x}. \]
Answer
\[ \lim_{x\to+\infty}x^{\frac1x}=1. \]
Solution
The base tends to \(+\infty\), while the exponent tends to zero: we are in the form \(\infty^0\). Set \(y(x)=x^{1/x}\). For \(x>0\),
\[ \ln y(x)=\frac{\ln x}{x}. \]
The logarithm grows more slowly than the power function \(x\); in symbols, \(\ln x=o(x)\) as \(x\to+\infty\). Hence
\[ \lim_{x\to+\infty}\frac{\ln x}{x}=0. \]
It follows that \(\ln y(x)\to0\) and, applying the exponential function,
\[ \lim_{x\to+\infty}x^{\frac1x}=e^0=1. \]
Exercise 19 — level ★★★★☆
Evaluate the limit
\[ \lim_{x\to+\infty}x^{-\frac1{\sqrt{\ln x}}}. \]
Answer
\[ \lim_{x\to+\infty}x^{-\frac1{\sqrt{\ln x}}}=0. \]
Solution
As \(x\to+\infty\), the base tends to \(+\infty\), while \(\ln x\to+\infty\) and hence the exponent \(\displaystyle-\frac1{\sqrt{\ln x}}\) tends to zero. The limit presents the form \(\infty^0\).
Set \(y(x)=x^{-1/\sqrt{\ln x}}\). For \(x>1\), the function is positive and we may write
\[ \ln y(x)=-\frac1{\sqrt{\ln x}}\ln x=-\sqrt{\ln x}. \]
Since \(\ln x\to+\infty\), we also have \(\sqrt{\ln x}\to+\infty\). Consequently,
\[ \ln y(x)\to-\infty. \]
Applying the exponential function, we obtain
\[ y(x)=e^{\ln y(x)}=e^{-\sqrt{\ln x}}\to0. \]
This exercise shows that a form of type \(\infty^0\) does not necessarily yield \(1\): the result depends on the product of the exponent and the logarithm of the base.
Exercise 20 — level ★★★★★
Let \(a,b\in\mathbb{R}\). Evaluate the limit
\[ \lim_{x\to0}\left(\frac{1+ax}{1+bx}\right)^{\frac1x}. \]
Answer
\[ \lim_{x\to0}\left(\frac{1+ax}{1+bx}\right)^{\frac1x}=e^{a-b}. \]
Solution
As \(x\to0\), both \(1+ax\) and \(1+bx\) tend to \(1\), so the base tends to \(1\). The exponent \(\displaystyle\frac1x\) diverges in absolute value. The limit therefore presents the form \(1^\infty\).
For \(x\) sufficiently close to zero, both factors \(1+ax\) and \(1+bx\) are positive; their ratio is therefore likewise positive. Set
\[ y(x)=\left(\frac{1+ax}{1+bx}\right)^{\frac1x}. \]
Taking the logarithm and using the property of the logarithm of a quotient, we obtain
\[ \ln y(x)=\frac1x\left[\ln(1+ax)-\ln(1+bx)\right]. \]
We separate the two terms:
\[ \ln y(x)=\frac{\ln(1+ax)}x-\frac{\ln(1+bx)}x. \]
If \(a\neq0\), we may write
\[ \frac{\ln(1+ax)}x=a\frac{\ln(1+ax)}{ax}\to a, \]
since \(ax\to0\) and \(\displaystyle\frac{\ln(1+t)}t\to1\). If \(a=0\), the term is identically zero and likewise tends to \(a=0\). By the same reasoning,
\[ \frac{\ln(1+bx)}x\to b. \]
Hence
\[ \lim_{x\to0}\ln y(x)=a-b. \]
The exponential function is continuous, so
\[ \lim_{x\to0}y(x)=e^{a-b}. \]
The formula also encompasses some noteworthy special cases: if \(a=b\), the base is identically equal to \(1\) and the limit equals \(1=e^0\); if \(a>b\), the limit exceeds \(1\); if \(a<b\), it lies between \(0\) and \(1\).