L'Hospital Rule

There are many techniques to calculate the limit of a function. But do you know that there exists a derivative method also to evaluate the limiting value of a function. L'Hospital's rule is used in calculus to determine the limits of indeterminate forms of the type $\frac{0}{0},\frac{\infty }{\infty }$ using derivatives.

L’Hospital’s rule many times helps in solving complex limit problems in an easier way.

Table of contents:

• L'Hospital’s Rule
• Practice problems
• FAQs

L'Hospital’s Rule

L'Hospital's rule states that for functions f(x)andg(x) which are differentiable on an open interval I except possibly at a point c contained in I such that

If $lim\backslash \mathrm{below}x\to a\frac{f\left(x\right)}{g\left(x\right)}$ takes $\frac{0}{0},\frac{\infty }{\infty }$ or form,then

$lim\backslash \mathrm{below}x\to a\frac{f\left(x\right)}{g\left(x\right)}=lim\backslash \mathrm{below}x\to a\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}$

This rule can be applied multiple times while evaluating the limit. The key concept is to differentiate the numerator and the denominator separately until the indeterminate form gets eliminated.

Note: For indeterminate forms other than $\frac{0}{0},\frac{\infty }{\infty }$, , L’Hospital rule can be applied only after changing it to either 00 or form.

Example: Evaluate

Here first we will check the indeterminate form of the given expression by putting x=0

$\frac{sin 0}{0}=\frac{0}{0}$ which clearly means that we can apply the L'Hospital Rule.

Now differentiating the numerator and denominator independently, we get

$lim\backslash \mathrm{below}x\to 0\frac{cos x}{1}$ (Now, the indeterminate form is removed)

Hence, now directly we can evaluate the limiting value in the next step i.e.

$lim\backslash \mathrm{below}x\to 0\frac{cos 0}{1}=\frac{1}{1}=1$

Related Concept Video: L'Hospital Rule | Limits | Class 11 & 12 MATHS | JEE 2021/22 | JEE 2023 l Keshav Sir | BYJU'S JEE

Practice problems:

Example: Evaluate $lim\backslash \mathrm{below}x\to 2\frac{sin(e^{x-2}-1)}{ln(x-1)}$

Solution:

$lim\backslash \mathrm{below}x\to 2\frac{sin(e^{x-2}-1)}{ln(x-1)}\left(\frac{0}{0}form\right)$

Applying L'Hospital Rule

$\Rightarrow lim\backslash \mathrm{below}x\to 2\frac{cos (e^{x-2}-1)\times e^{x-2}}{\frac{1}{(x-1)}}$

$\Rightarrow lim\backslash \mathrm{below}x\to 2cos (e^{x-2}-1)\times e^{x-2}\times (x-1)$

$\Rightarrow lim\backslash \mathrm{below}x\to 2cos (e^{2-2}-1)\times e^{2-2}\times (2-1)$

$\Rightarrow lim\backslash \mathrm{below}x\to 2cos \left(0\right)\times e^0\times \left(1\right)=1$

$\therefore lim\backslash \mathrm{below}x\to 2\frac{sin(e^{x-2}-1)}{ln(x-1)}=1$

Example: If $f\left(9\right)=4andf\text{'}\left(9\right)=3thenlim\backslash \mathrm{below}x\to 9\frac{\sqrt{f\left(x\right)}-2}{\sqrt{x}-3}.$

Solution:

$lim\backslash \mathrm{below}x\to 9\frac{\sqrt{f\left(x\right)}-2}{\sqrt{x}-3}(\frac{0}{0} form)$

Now applying L'Hospital Rule

$\Rightarrow lim\backslash \mathrm{below}x\to 9\frac{\frac{1}{2\sqrt{f\left(x\right)}}f\text{'}\left(x\right)}{\frac{1}{2\sqrt{x}}}$

$$\begin{gathered} \Rightarrow lim\backslash \mathrm{below}x\to 9\frac{1}{2\sqrt{f\left(x\right)}}f\text{'} \\ \left(x\right)\times 2\sqrt{x}\Rightarrow lim\backslash \mathrm{below}x\to 9\frac{1}{2\sqrt{f\left(9\right)}} \\ f\text{'}\left(9\right)\times 2\sqrt{9}\Rightarrow \frac{1}{2\sqrt{4}}\times 3\times 2\sqrt{9} \end{gathered}$$

$\Rightarrow \frac{9}{2}$

$\therefore lim\backslash \mathrm{below}x\to 9\frac{\sqrt{f\left(x\right)}-2}{\sqrt{x}-3}= \frac{9}{2}$

Example: If $f\left(a\right)=2, f\text{'}\left(a\right)=1, g\left(a\right)=-1; g\text{'}\left(a\right)=2,thenlim\backslash \mathrm{below}x\to a\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}.$

Solution:

$lim\backslash \mathrm{below}x\to a\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}(\frac{0}{0} form)$

Applying L'Hospital Rule

$$\begin{gathered} \Rightarrow lim\backslash \mathrm{below}x\to a\frac{g\text{'}\left(x\right)f\left(a\right)-g\left(a\right)f\text{'}\left(x\right)}{1} \\ \Rightarrow lim\backslash \mathrm{below}x\to a\frac{g\text{'}\left(a\right)f\left(a\right)-g\left(a\right)f\text{'}\left(a\right)}{1}=\frac{2\times 2-(-1)\times 1}{1}=5 \\ \therefore lim\backslash \mathrm{below}x\to a\frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}=5 \end{gathered}$$

Example: If the graph of the function y=f(x) has a unique tangent at the point (a,0) through which the graph passes,then evaluate

Solution:

The graph passes through point (a,0) f(a)=0

Also, f(x) is differentiable at x=a ( Since, it has unique tangent at x=a )

Now, is of 00 form.Therefore, L'Hospital Rule can be applied

Example:$lim\backslash \mathrm{below}x\to \infty \frac{3x^2-4x+3}{5x-7}$

Solution:

$lim\backslash \mathrm{below}x\to \infty \frac{3x^2-4x+3}{5x-7}(\frac{\infty }{\infty } form)$

Applying L'Hospital Rule

$$\begin{gathered} \Rightarrow lim\backslash \mathrm{below}x\to \infty \frac{6x^{-}4}{5}\Rightarrow \frac{1}{5} \\ lim\backslash \mathrm{below}x\to \infty 6x-4 =\infty \\ \therefore lim\backslash \mathrm{below}x\to \infty \frac{3x^2-4x+3}{5x-7}=\infty \left(Notdefined\right) \end{gathered}$$

Example:$:lim\backslash \mathrm{below}x\to \infty \frac{3x^{-}4}{5x^2-7}$

Solution:

$lim\backslash \mathrm{below}x\to \infty \frac{3x^{-}4}{5x^2-7}(\frac{\infty }{\infty } form)$

Applying L'Hospital Rule

$$\begin{gathered} \Rightarrow lim\backslash \mathrm{below}x\to \infty \frac{3}{10x}=\frac{3}{10\times \infty }=0 \\ \therefore lim\backslash \mathrm{below}x\to \infty \frac{3x^{-}4}{5x^2-7}=0 \end{gathered}$$

Example: Evaluate$lim\backslash \mathrm{below}n\to \infty \frac{\sum n^2}{n^3}.$

Solution:

$lim\backslash \mathrm{below}n\to \infty \frac{\sum n^2}{n^3}(\frac{\infty }{\infty } form)$

$$\begin{gathered} Weknow\sum n^2=\frac{n(n+1)(2n+6)}{6} \\ \Rightarrow lim\backslash \mathrm{below}n\to \infty \frac{n(n+1)(2n+6)}{{6n}^3} \\ \Rightarrow lim\backslash \mathrm{below}n\to \infty \frac{(n+1)(2n+6)}{{6n}^2}(\frac{\infty }{\infty } form) \end{gathered}$$

FAQs

Question.1 Why does L'Hopital's rule not always work?

Answer: The L’ Hospital's Rule occasionally fails by becoming trapped in an endless loop. For example we might encounter a situation in which after applying the L’Hospital rule multiple times, the limit returns to the original limit, indicating that we will never reach to a conclusion.

Question.2 Can we use the standard formulas of limits and L'Hospital Rule simultaneously?

Answer: Yes the standard limit formulas can be used along with the L’Hospital Rule

Question.3 Why is this rule named L' Hospital?

Answer: The rule derives its name from a French mathematician that it why it has been named so.