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.

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 $l i m \ below x \to a \frac{f (x)}{g (x)}$ takes $\frac{0}{0}, \frac{\infty}{\infty}$ or form,then

$l i m \ below x \to a \frac{f (x)}{g (x)} = l i m \ below x \to a \frac{f' (x)}{g' (x)}$

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

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

Now differentiating the numerator and denominator independently, we get

$l i m \ below x \to 0 \frac{c o s x}{1}$ (Now, the indeterminate form is removed)

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

$l i m \ below x \to 0 \frac{c o s 0}{1} = \frac{1}{1} = 1$

Related Concept Video: L'Hospital Rule | Limits | Class 11 & 12 MATHS

Practice problems:

Example: Evaluate $l i m \ below x \to 2 \frac{s i n (e^{x - 2} - 1)}{l n (x - 1)}$

Solution:

$l i m \ below x \to 2 \frac{s i n (e^{x - 2} - 1)}{l n (x - 1)} (\frac{0}{0} f o r m)$

Applying L'Hospital Rule

$\Rightarrow l i m \ below x \to 2 \frac{c o s (e^{x - 2} - 1) \times e^{x - 2}}{\frac{1}{(x - 1)}}$

$\Rightarrow l i m \ below x \to 2 c o s (e^{x - 2} - 1) \times e^{x - 2} \times (x - 1)$

$\Rightarrow l i m \ below x \to 2 c o s (e^{2 - 2} - 1) \times e^{2 - 2} \times (2 - 1)$

$\Rightarrow l i m \ below x \to 2 c o s (0) \times e^{0} \times (1) = 1$

$∴ l i m \ below x \to 2 \frac{s i n (e^{x - 2} - 1)}{l n (x - 1)} = 1$

Example: If $f (9) = 4 a n d f' (9) = 3 t h e n l i m \ below x \to 9 \frac{\sqrt{f (x)} - 2}{\sqrt{x} - 3} .$

Solution:

$l i m \ below x \to 9 \frac{\sqrt{f (x)} - 2}{\sqrt{x} - 3} (\frac{0}{0} f o r m)$

Now applying L'Hospital Rule

$\Rightarrow l i m \ below x \to 9 \frac{\frac{1}{2 \sqrt{f (x)}} f' (x)}{\frac{1}{2 \sqrt{x}}}$

$\Rightarrow l i m \ below x \to 9 \frac{1}{2 \sqrt{f (x)}} f' (x) \times 2 \sqrt{x} \Rightarrow l i m \ below x \to 9 \frac{1}{2 \sqrt{f (9)}} f' (9) \times 2 \sqrt{9} \Rightarrow \frac{1}{2 \sqrt{4}} \times 3 \times 2 \sqrt{9}$

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

$∴ l i m \ below x \to 9 \frac{\sqrt{f (x)} - 2}{\sqrt{x} - 3} = \frac{9}{2}$

Example: If $f (a) = 2, f' (a) = 1, g (a) = - 1; g' (a) = 2, t h e n l i m \ below x \to a \frac{g (x) f (a) - g (a) f (x)}{x - a} .$

Solution:

$l i m \ below x \to a \frac{g (x) f (a) - g (a) f (x)}{x - a} (\frac{0}{0} f o r m)$

Applying L'Hospital Rule

$\Rightarrow l i m \ below x \to a \frac{g' (x) f (a) - g (a) f' (x)}{1} \Rightarrow l i m \ below x \to a \frac{g' (a) f (a) - g (a) f' (a)}{1} = \frac{2 \times 2 - (- 1) \times 1}{1} = 5 ∴ l i m \ below x \to a \frac{g (x) f (a) - g (a) f (x)}{x - a} = 5$

Example: $l i m \ below x \to \infty \frac{3 x^{2} - 4 x + 3}{5 x - 7}$

Solution:

$l i m \ below x \to \infty \frac{3 x^{2} - 4 x + 3}{5 x - 7} (\frac{\infty}{\infty} f o r m)$

Applying L'Hospital Rule

$\Rightarrow l i m \ below x \to \infty \frac{6 x^{-} 4}{5} \Rightarrow \frac{1}{5} l i m \ below x \to \infty 6 x - 4 = \infty ∴ l i m \ below x \to \infty \frac{3 x^{2} - 4 x + 3}{5 x - 7} = \infty (N o t d e f i n e d)$

Example: $: l i m \ below x \to \infty \frac{3 x^{-} 4}{5 x^{2} - 7}$

Solution:

$l i m \ below x \to \infty \frac{3 x^{-} 4}{5 x^{2} - 7} (\frac{\infty}{\infty} f o r m)$

Applying L'Hospital Rule

$\Rightarrow l i m \ below x \to \infty \frac{3}{10 x} = \frac{3}{10 \times \infty} = 0 ∴ l i m \ below x \to \infty \frac{3 x^{-} 4}{5 x^{2} - 7} = 0$

Example: Evaluate $l i m \ below n \to \infty \frac{\sum n^{2}}{n^{3}} .$

Solution:

$l i m \ below n \to \infty \frac{\sum n^{2}}{n^{3}} (\frac{\infty}{\infty} f o r m)$

$W e k n o w \sum n^{2} = \frac{n (n + 1) (2 n + 6)}{6} \Rightarrow l i m \ below n \to \infty \frac{n (n + 1) (2 n + 6)}{{6 n}^{3}} \Rightarrow l i m \ below n \to \infty \frac{(n + 1) (2 n + 6)}{{6 n}^{2}} (\frac{\infty}{\infty} f o r m)$

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.