L'Hospital's Rule in Calculus: Formula, Proof and Example

Limits are one of the most basic ideas in calculus, where one can learn how functions behave as they approach particular points. Of interest, though, is that some limits tend not to be as straightforward as finding the others, such that they could evaluate indeterminate forms. An indeterminate form in mathematics means that we cannot be able to determine the original value even after the substitution of the limits.

In this article, we will explore the concept of L'Hospital's Rule, an essential topic in Class 11 Mathematics under Calculus. This concept is vital not only for board exams but also for various competitive exams, including the Joint Entrance Examination (JEE) Main, SRM Joint Engineering Entrance, BITSAT, WBJEE, and BCECE. Between 2013 and 2023, a total of ten questions on this topic were featured in JEE Main: one question in 2016, one in 2018, three in 2019, one in 2020, three in 2021, and one in 2022.

L’Hospital’s Rule

L'Hospital's rule is a general method of evaluating indeterminate forms such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. To evaluate the limits of indeterminate forms for the derivatives in calculus, L'Hospital's rule is used. You can apply this rule still it holds any indefinite form every time after its application.

L'Hospital's Rule states that, if $\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}$ is of $\frac{0}{0}$ or $\frac{\infty}{\infty}$ form , then differentiate numerator and denominator till this intermediate form is removed. $\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$

But, if we again get the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then, $\lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime \prime}(x)}{g^{\prime \prime}(x)}$ (so we differentiate numerator and denominator again)

This process is continued till the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$ is removed.

Note:

We do not use the quotient rule of differentiation here. The numerator and denominator have to be differentiated separately.

Some Application of L’Hospital’s Rule.

(i) $\lim\limits _{x \rightarrow 0} \frac{\sin x}{x}=\lim\limits _{x \rightarrow 0} \frac{\cos x}{1}=1$
(ii) $\lim\limits _{x \rightarrow \infty} \frac{\log _e x}{x}=\lim\limits _{x \rightarrow \infty} \frac{1 / x}{1}=0$
(iii) $\lim\limits _{x \rightarrow 0} \frac{\log _e(1+x)}{x}=\lim\limits _{x \rightarrow 0} \frac{\frac{1}{1+x}}{1}=1$

Note:

In some cases, L’Hospital’s Rule fails to evaluate the limit:

For example,

$\begin{equation}
\begin{aligned}
&\begin{aligned}
\lim\limits _{x \rightarrow \infty} & \frac{x+\cos x}{x-\sin x} \quad\left(\frac{\infty}{\infty} \text { form }\right) \\
& =\lim\limits _{x \rightarrow \infty} \frac{1-\sin x}{1-\cos x}, \text { which is cannot be calculated }
\end{aligned}\\
&\text { The correct value of this limit is }\\
&\lim\limits _{x \rightarrow \infty} \frac{x+\cos x}{x-\sin x}=\lim\limits _{x \rightarrow \infty} \frac{1+\frac{\cos x}{x}}{1-\frac{\sin x}{x}}=\frac{1+0}{1-0}=1
\end{aligned}
\end{equation}$

Let’s go through the illustration to understand how to solve such type of questions

$\begin{equation}
\text { The value of the limit } \lim\limits _{x \rightarrow \infty}(\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2}) \text { is }
\end{equation}$

First Rationalising the expression

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& =\lim\limits _{x \rightarrow \infty} \frac{(\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2})(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{(x+2 a)(2 x+a)-2 x^2}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{2 x^2+5 a x+2 a^2-2 x^2}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{5 a x+2 a^2}{\sqrt{2 x^2+5 a x+2 a^2}+x \sqrt{2}}
\end{aligned}\\
&\text { Dividing numerator and denominator by } \mathrm{x} \text {, we get : }\\
&=\lim\limits _{x \rightarrow \infty} \frac{5 a+\frac{2 a^2}{x}}{\sqrt{2+\frac{5 a}{x}+\frac{2 a^2}{x^2}}+\sqrt{2}}=\frac{5 a}{2 \sqrt{2}}
\end{aligned}
\end{equation}$

Solved Examples Based on L'Hopital's Rule

Example 1: If $f(x)$ is a differentiable function in the interval $(0, \infty)$ such that $f(1)=1$ and $\lim\limits _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$, for each $x>0$, then $f(3 / 2)$ is equal to : [JEE Main 2016]
1) $\frac{13}{6}$
2) $\frac{23}{18}$
3) $\frac{25}{9}$
4) $\frac{31}{18}$

Solution:

As we learned in L - Hospital Rule -

$\begin{equation}
\begin{aligned}
&\text { In the form of } \frac{0}{0} \text { and } \frac{\infty}{\infty} \text { we dif ferentiate } \frac{N^r}{D^r} \text { separately. }\\
&\Rightarrow \lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}
\end{aligned}
\end{equation}$

- wherein

$\begin{equation}
\begin{aligned}
&\lim\limits _{x \rightarrow a} \frac{\frac{d}{d x} f(x)}{\frac{d}{d x} g(x)}\\
&\text { Where } f(x) \text { and } g(x)=0\\
&\begin{aligned}
& \lim\limits _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1 \\
& \lim\limits _{t \rightarrow x} \frac{2 t f(x)-x^2 f^{\prime}(t)}{1}=1 \\
& \therefore 2 x f(x)-x^2 f^{\prime}(x)=1
\end{aligned}
\end{aligned}
\end{equation}$

$\begin{equation}
\begin{aligned}
& \text { Now, let } \mathrm{y}=\mathrm{f}(\mathrm{x}) \\
& 2 x y-x^2 \frac{d y}{d x}=1 \\
& \Rightarrow x^2 \frac{d y}{d x}-2 x y=-1 \\
& \Rightarrow \frac{d y}{d x}-\frac{2}{x} y=-\frac{1}{x^2} \\
& P=-\frac{2}{x} \text { and } Q=-\frac{1}{x^2} \\
& \therefore \int P d x=-2 \int \frac{d x}{x}=-2 \log x=\log \frac{1}{x^2} \\
& \therefore e^{\log \frac{1}{x^2}}=\frac{1}{x^2}
\end{aligned}
\end{equation}$

$\therefore$ Solution is

$
\begin{aligned}
& y \cdot \frac{1}{x^2}=\int-\frac{1}{x^2} \times \frac{1}{x^2} d x=\int \frac{1}{x^4} d x \\
& =-\int x^{-4} d x=\frac{-x^{-4+1}}{-4+1}+C \\
& \frac{y}{x^2}=\frac{-x^{-3}}{-3}+C=C+\frac{1}{3 x^3}
\end{aligned}
$

Put, $x=1, y=1$

$
\begin{aligned}
& \frac{1}{1}=C+\frac{1}{3} \\
& \therefore C=1-\frac{1}{3}=\frac{2}{3} \\
& \frac{y}{x^2}=\frac{1}{3 x^3}+\frac{2}{3} \\
& \therefore y=\frac{1}{3 x}+\frac{2 x^2}{3}
\end{aligned}
$

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \text { Put, } x=\frac{3}{2} \\
& y=\frac{1}{3 \times \frac{3}{2}}+\frac{2}{3} \times \frac{9}{4} \\
& =\frac{2}{9}+\frac{3}{2}=\frac{4+27}{18}=\frac{31}{18}
\end{aligned}\\
&\text { Hence, the answer is the option } 4 .
\end{aligned}
\end{equation}$

Example 2: If the function f is defined as $f(x)=\frac{1}{x}-\frac{k-1}{e^{2 x}-1}$ , $x \neq 0$ and is continuous at $\mathrm{x}=0$, then the ordered pair $(\mathrm{k}, \mathrm{f}(0))$ is equal to: [JEE Main 2018]
1) $(3,2)$
2) $(3,1)$
3) $(2,1)$
4) $(1 / 3,2)$

Solution:

As we have learned to calculate indeterminate limits:

Now, $\lim\limits _{x \rightarrow 0} \frac{1}{x}-\frac{k-1}{e^{2 x}-1}$

$
\lim\limits _{x \rightarrow 0} \frac{\left(\left(e^{2 x}-1\right)-x(k-1)\right)}{x\left(e^{2 x}-1\right)}
$

By using the L' Hospital's rule

$
\begin{aligned}
& \frac{2 e^{2 x}-1(k-1)}{e^{2 x}-1+2 x e^{2 x}} \\
& \text { put } \mathrm{k}=3 \\
& \mathrm{f}(0)=1
\end{aligned}
$

Hence, the answer is the option 2.

Example 3: $\lim\limits _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^2+2 \sin x+1}-\sqrt{\sin ^2 x-x+1}}$ is:
[JEE Main 2019]
1) 2
2) 6
3) 3
4) 1

Solution:

L - Hospital Rule -

In the form of $\frac{0}{0}$ and $\frac{\infty}{\infty}$ we dif ferentiate $\frac{N^r}{D^r}$ separately.
$\Rightarrow \lim\limits _{x \rightarrow a} \frac{f(x)}{g(x)}=\lim\limits _{x \rightarrow a} \frac{f^{\prime}(x)}{g^{\prime}(x)}$ Where $f(x)$ and $g(x)=0$

$\begin{equation}
\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^2+2 \sin x+1}-\sqrt{\sin ^2 x-x+1}} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{\sin ^2 x-x+1}\right)}{\left(\sqrt{x^2+2 \sin x+1}\right)^2-\left(\sqrt{\sin ^2 x-x+1}\right)^2} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{1-\cos ^2 x-x+1}\right)}{\left(x^2+2 \sin x+1\right)-\left(\sin ^2 x-x+1\right)} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{2-\cos ^2 x-x}\right)}{x^2+2 \sin x-\sin ^2 x+x} \\
& \Rightarrow \lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)(2)}{x^2+2 \sin x-\sin ^2 x+x}
\end{aligned}
\end{equation}$

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \frac{0}{0} \text { form use L'Hopital rule } \\
& +\lim\limits _{x \rightarrow 0} \frac{(1+2 \cos x) \times 2}{2 x+2 \cos x-2 \sin x \cos x+1} \\
& =>\frac{(1+2)(2)}{0+1+2-0} \\
& =>2
\end{aligned}\\
&\text { Hence, the answer is the option } 1 .
\end{aligned}
\end{equation}$

Example 4: Let $f:(0, \infty) \rightarrow(0, \infty)$ be a differentiable function such that $\mathrm{f}(1)=\mathrm{e}$ and $\lim\limits _{t \rightarrow x} \frac{t^2 f^2(x)-x^2 f^2(t)}{t-x}=0$. If $\mathrm{f}(\mathrm{x})=1$, then x is equal to: [JEE Main 2020]
1) $\frac{1}{e}$
2) 2 e
3) $\frac{1}{2 e}$
4) e

Solution:

$\begin{equation}
\begin{aligned}
&\begin{aligned}
& L=\operatorname{Lim}_{t \rightarrow x} \frac{t^2 f^2(x)-x^2 f^2(t)}{t-x} \\
& \text { using } \mathrm{L} \text { H'opital. rule } \\
& \mathrm{L}=\operatorname{Lim}_{\mathrm{t} \rightarrow \mathrm{x}} \frac{2 \mathrm{tf}^2(\mathrm{x})-\mathrm{x}^2 \cdot 2 \mathrm{f}^{\prime}(\mathrm{t}) \cdot \mathrm{f}(\mathrm{t})}{1} \\
& \Rightarrow \mathrm{L}=2 \mathrm{xf}(\mathrm{x})\left(\mathrm{f}(\mathrm{x})-\mathrm{xf}^{\prime}(\mathrm{x})\right)=0(\text { given }) \\
& \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{xf}^{\prime}(\mathrm{x}) \Rightarrow \int \frac{\mathrm{f}^{\prime}(\mathrm{x}) \mathrm{dx}}{\mathrm{f}(\mathrm{x})}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\
& \Rightarrow \ln l \mathrm{f}(\mathrm{x})|=\ln | \mathrm{x} \mid+\mathrm{C} \\
& \because \mathrm{f}(1)=\mathrm{e}, \mathrm{x}>0, \mathrm{f}(\mathrm{x})>0 \\
& \Rightarrow \mathrm{f}(\mathrm{x})=\operatorname{ex}, \text { if } \mathrm{f}(\mathrm{x})=1 \Rightarrow \mathrm{x}=\frac{1}{\mathrm{e}}
\end{aligned}\\
&\text { Hence, the answer is option (1). }
\end{aligned}
\end{equation}$

Example 5: If $x \rightarrow 1 \frac{x^2-a x+b}{x-1}=5$, then a+b is equal to:
[JEE Main 2019]
1) -7
2) 5
3) -4
4) 1

Solution:

$\lim\limits _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}=5
$

As $x \rightarrow 1$ the denominator will become 0
For a finite limit(=5 in this case) numerator must also approach zero as $x \rightarrow 1$.

So, $1-a+b=0$.
Now,

$
\lim\limits _{x \rightarrow 1} \frac{x^2-a x+b}{x-1}\left(\frac{0}{0} \text { form }\right)
$

Applying L-Hospital's Rule

$
\begin{aligned}
& \lim\limits _{x \rightarrow 1} \frac{2 x-a}{1}=5 \\
& 2-a=5 \\
& \Rightarrow a=-3
\end{aligned}
$

and from (i)

$
b=a-1 \Rightarrow b=-3-1=-4
$

So, $a+b=-3-4=-7$
Hence, the answer is the option (1).

Summary: Indeterminate forms can be evaluated by applying appropriate transformations—L'Hôpital's Rule, factorization, and logarithmic manipulation. It is used to evaluate the limits of indeterminate forms. Calculus was created to describe how the quantities change. The concept of limit is the cornerstone on which the development of calculus rests.