Imagine you are filling a swimming pool with water using two taps, and you want to know how the water level changes at the exact moment both taps are adjusted together. Calculating the exact rate of change can sometimes lead to tricky situations where both the numerator and denominator in a rate formula approach zero. This is where L’Hospital’s Rule comes in handy. In mathematics, L’Hospital’s Rule helps solve indeterminate forms like $0/0$ and $\infty/\infty$, making it easier to evaluate limits of functions that seem impossible to solve at first glance. In this article, we will cover the formula of L’Hospital’s Rule, step-by-step examples, common applications, and practice questions.
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L’Hospital’s Rule is a key tool in calculus used to evaluate limits of functions that are initially indeterminate, such as $0/0$ or $\infty/\infty$. It uses derivatives of the numerator and denominator to simplify complex limits.
Indeterminate forms appear frequently in problems involving exponential, logarithmic, trigonometric functions, and rational functions. Mastering L’Hospital’s Rule allows solving these tricky limits accurately and efficiently.
If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or $\pm \infty$, and $f'(x)$ and $g'(x)$ exist near $x = a$, then $\lim_{x \to a} f(x)/g(x) = \lim_{x \to a} f'(x)/g'(x)$ provided the right-hand side exists or is $\pm \infty$.
Conditions for Applying the Rule:
The limit must be in the form $0/0$ or $\infty/\infty$.
Both numerator $f(x)$ and denominator $g(x)$ must be differentiable near $x = a$.
$g'(x)$ must not be zero near $x = a$.
If the limit remains indeterminate, the rule can be applied repeatedly using higher-order derivatives.
1. $0/0$ Form
Example: (i) $\lim\limits_{x \to 0} \sin x / x = \lim\limits_{x \to 0} \cos x / 1 = 1$
2. $\infty/\infty$ Form
Example: (ii) $\lim\limits_{x \to \infty} \log_e x / x = \lim\limits_{x \to \infty} 1/x / 1 = 0$
3. Other Derived Forms (can be converted to $0/0$ or $\infty/\infty$)
$0 \cdot \infty$: convert to fraction $f(x) \cdot g(x) = f(x)/(1/g(x))$
$1^\infty$: use logarithms $\lim\limits_{x \to a} [1 + f(x)]^{g(x)} = e^{\lim\limits_{x \to a} f(x) g(x)}$
$0^0$: use logarithms $\lim\limits_{x \to a} f(x)^{g(x)} = e^{\lim\limits_{x \to a} g(x) \ln f(x)}$
$\infty^0$: use logarithms $\lim\limits_{x \to a} f(x)^{g(x)} = e^{\lim\limits_{x \to a} g(x) \ln f(x)}$
4. Logarithmic limit example: (iii) $\lim\limits_{x \to 0} \log_e (1+x) / x = \lim\limits_{x \to 0} 1/(1+x) / 1 = 1$
Note:
We do not use the quotient rule of differentiation here. The numerator and denominator have to be differentiated separately.
This table provides a quick reference for all common indeterminate forms like $0/0$, $\infty/\infty$, $0 \cdot \infty$, and others. It helps students identify the type of limit and the appropriate method, including L’Hospital’s Rule, for solving them efficiently.
| S. No | Indeterminate Form | Description |
|---|---|---|
| 1 | $0/0$ | Standard L’Hospital’s Rule can be applied directly. |
| 2 | $\infty/\infty$ | Standard L’Hospital’s Rule can be applied directly. |
| 3 | $0 \cdot \infty$ | Convert to fraction: $f(x) \cdot g(x) = f(x) / (1/g(x))$ and apply L’Hospital’s Rule. |
| 4 | $1^\infty$ | Use logarithms: $\lim\limits_{x \to a} [1+f(x)]^{g(x)} = e^{\lim\limits_{x \to a} f(x) g(x)}$ |
| 5 | $0^0$ | Use logarithms: $\lim\limits_{x \to a} f(x)^{g(x)} = e^{\lim\limits_{x \to a} g(x) \ln f(x)}$ |
| 6 | $\infty^0$ | Use logarithms: $\lim\limits_{x \to a} f(x)^{g(x)} = e^{\lim\limits_{x \to a} g(x) \ln f(x)}$ |
| 7 | $\infty - \infty$ | Convert to a fraction or common expression, then apply L’Hospital’s Rule. |
The Formula of L’Hospital’s Rule provides a simple and powerful way to evaluate limits that are initially indeterminate, such as $0/0$ or $\infty/\infty$. By using derivatives of the numerator and denominator, complex limits can be simplified into solvable forms, making it essential for Class 11, Class 12, JEE, NEET, and other competitive exams.
(i) $\lim\limits_{x \to a} \frac{f(x)}{g(x)} = \lim\limits_{x \to a} \frac{f'(x)}{g'(x)}$, if $\lim\limits_{x \to a} f(x) = \lim\limits_{x \to a} g(x) = 0$ or $\pm \infty$
(ii) Repeated application for higher-order derivatives:
$\lim\limits_{x \to 0} \frac{1 - \cos x}{x^2} = \lim\limits_{x \to 0} \frac{\sin x}{2x} = \lim\limits_{x \to 0} \frac{\cos x}{2} = 1/2$
(i) Verify indeterminate form before applying, e.g., check if it is $0/0$ or $\infty/\infty$
(ii) Check differentiability of numerator and denominator near the point of interest
(iii) Simplify expressions when possible before applying:
$\lim\limits_{x \to 0} \frac{\sin^2 x}{x^2} = \lim\limits_{x \to 0} (\sin x / x)^2 = 1$
(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
$\begin{equation}
\begin{aligned}
&\begin{aligned}
& =\lim\limits _{x \rightarrow \infty} \frac{(\sqrt{(x+2 a)(2 x+a)}-x \sqrt{2})(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{(x+2 a)(2 x+a)-2 x^2}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{2 x^2+5 a x+2 a^2-2 x^2}{(\sqrt{(x+2 a)(2 x+a)}+x \sqrt{2})} \\
& =\lim\limits _{x \rightarrow \infty} \frac{5 a x+2 a^2}{\sqrt{2 x^2+5 a x+2 a^2}+x \sqrt{2}}
\end{aligned}\\
&\text { Dividing numerator and denominator by } \mathrm{x} \text {, we get : }\\
&=\lim\limits _{x \rightarrow \infty} \frac{5 a+\frac{2 a^2}{x}}{\sqrt{2+\frac{5 a}{x}+\frac{2 a^2}{x^2}}+\sqrt{2}}=\frac{5 a}{2 \sqrt{2}}
\end{aligned}
\end{equation}$
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 $g(x)=0$
$\begin{equation}
\begin{aligned}
& \lim\limits _{x \rightarrow 0} \frac{x+2 \sin x}{\sqrt{x^2+2 \sin x+1}-\sqrt{\sin ^2 x-x+1}} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{\sin ^2 x-x+1}\right)}{\left(\sqrt{x^2+2 \sin x+1}\right)^2-\left(\sqrt{\sin ^2 x-x+1}\right)^2} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{1-\cos ^2 x-x+1}\right)}{\left(x^2+2 \sin x+1\right)-\left(\sin ^2 x-x+1\right)} \\
& =>\lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)\left(\sqrt{x^2+2 \sin x+1}+\sqrt{2-\cos ^2 x-x}\right)}{x^2+2 \sin x-\sin ^2 x+x} \\
& \Rightarrow \lim\limits _{x \rightarrow 0} \frac{(x+2 \sin x)(2)}{x^2+2 \sin x-\sin ^2 x+x}
\end{aligned}
\end{equation}$
$\begin{equation}
\begin{aligned}
&\begin{aligned}
& \frac{0}{0} \text { form use L'Hospital rule } \\
& +\lim\limits _{x \rightarrow 0} \frac{(1+2 \cos x) \times 2}{2 x+2 \cos x-2 \sin x \cos x+1} \\
& =>\frac{(1+2)(2)}{0+1+2-0} \\
& =>2
\end{aligned}\\
&\text { Hence, the answer is the option } 1 .
\end{aligned}
\end{equation}$
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).
This section provides a comprehensive list of topics related to limits, including indeterminate forms, algebra of limits, expansion methods, left and right-hand limits, and higher-order derivatives, helping students structure their calculus preparation effectively for exams.
Here we offer a collection of NCERT notes, solutions, and exemplar problems for Limits and Derivatives, making it easier for students to follow the official syllabus, strengthen core concepts, and practice important examples for Class 11 and Class 12 exams.
NCERT Notes for Class 11 Maths Chapter 13 - Limits and Derivatives
NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives
NCERT Exempar Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives
This section features practice questions on L’Hospital’s Rule, designed to improve problem-solving speed, accuracy, and exam readiness for Class 11, Class 12, and competitive exams.
L-Hospitals Rule - Practice Question MCQ
We have shared the links below to practice questions on the related topics to limits:
Frequently Asked Questions (FAQs)
L’Hospital’s Rule is a calculus method used to evaluate limits that are indeterminate, such as $0/0$ or $\infty/\infty$, by taking the derivatives of the numerator and denominator.
Yes. If after the first application the limit still results in an indeterminate form, the rule can be applied multiple times using higher-order derivatives until a determinate limit is obtained.
Common indeterminate forms include $0/0$, $\infty/\infty$, $0 \cdot \infty$, $1^\infty$, $0^0$, $\infty^0$, and $\infty - \infty$.
Exponential limits like $\lim\limits_{x \to 0} \frac{a^x - 1}{x}$ are evaluated using the derivative of $a^x$, giving $\log_e a$.
Indeterminate forms in mathematics are specific values that occurs in conditions when the original value of the function cannot be determined even after applying the limits.
