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 $l i m \ below x \to a \frac{f (x)}{g (x)}$ 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 $<math xmlns="http://www.w3.org/1998/Math/MathML"><munder><mi>lim</mi><mrow><mi>x</mi><mo>→</mo><mn>0</mn></mrow></munder><mfrac><mrow><mi>sin</mi><mi>x</mi></mrow><mi>x</mi></mfrac></math>$

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.

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

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: 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

$<math xmlns="http://www.w3.org/1998/Math/MathML"><munder><mrow><mi>lim</mi><mo> </mo></mrow><mrow><mi>x</mi><mo>→</mo><mi>a</mi></mrow></munder><mfrac><mrow><mo> </mo><msub><mi>log</mi><mi>e</mi></msub><mfenced><mrow><mn>1</mn><mo>+</mo><mn>6</mn><mi>f</mi><mfenced><mi>x</mi></mfenced></mrow></mfenced></mrow><mrow><mn>3</mn><mi>f</mi><mfenced><mi>x</mi></mfenced></mrow></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><mspace linebreak="newline"/></math>$

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, $<math xmlns="http://www.w3.org/1998/Math/MathML"><munder><mrow><mi>lim</mi><mo> </mo></mrow><mrow><mi>x</mi><mo>→</mo><mi>a</mi></mrow></munder><mfrac><mrow><mo> </mo><msub><mi>log</mi><mi>e</mi></msub><mfenced><mrow><mn>1</mn><mo>+</mo><mn>6</mn><mi>f</mi><mfenced><mi>x</mi></mfenced></mrow></mfenced></mrow><mrow><mn>3</mn><mi>f</mi><mfenced><mi>x</mi></mfenced></mrow></mfrac><mspace linebreak="newline"/><mspace linebreak="newline"/><mspace linebreak="newline"/></math>$ is of 00 form.Therefore, L'Hospital Rule can be applied

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.