L’ Hospital’s Rule - Practice Questions & MCQ

L’Hospital’s Rule

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

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

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

Note:

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

Some Application of L’Hospital’s Rule.

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

Note:
In some cases, L'Hospital's Rule fails to evaluate limit
For example,
$\lim _{x \rightarrow \infty} \frac{x+\cos x}{x-\sin x} \quad\left(\frac{\infty}{\infty}\right.$ form $)$
$=\lim _{x \rightarrow \infty} \frac{1-\sin x}{1-\cos x}$, which is cannot be calculated
The correct value of this limit is

$
\lim _{x \rightarrow \infty} \frac{x+\cos x}{x-\sin x}=\lim _{x \rightarrow \infty} \frac{1+\frac{\cos x}{x}}{1-\frac{\sin x}{x}}=\frac{1+0}{1-0}=1
$