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L’ Hospital’s Rule - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • L’ Hospital’s Rule is considered one the most difficult concept.

  • 189 Questions around this concept.

Solve by difficulty

If f(1)=1,f'(1)=2\; then\;      is

If f(x) is a differentiable function in the interval (0, ∞) such that f(1) = 1 and  \lim_{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 :

Using the method of L’ Hospital, evaluate the limit \lim _{x \rightarrow \infty} \frac{x(4 x+1)^2}{(x+4)\left(x^2+6 x-1\right)}:

\mathrm{ \lim _{x \rightarrow \pi / 3}\left(\frac{\tan ^3 x-3 \tan x}{\cos \left(x+\frac{\pi}{6}\right)}\right) \text { is equal to }}

\mathrm{ If\, \, f(x)=\left(\frac{x-5+\frac{6}{x}}{x-3+\frac{2}{x}}\right), x \in \mathbb{R}-\{0,1,2\} \, \, then \lim _{x \rightarrow 2} f(x) \, \, equals}

\mathrm{ \lim _{x \rightarrow 0}\left\{\frac{e^{x^2}-\cos x}{x^2}\right\} \text { is equal to }}

\mathrm{ \lim _{x \rightarrow 0}\left\{\frac{\cos 4 x-1}{x}\right\} \text { equals } }

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\mathrm{ \lim _{x \rightarrow a}\left\{\frac{x^a-a^x}{x^x-a^a}\right\} \text { is equal to }}

\mathrm{ \text { If } f(x)=-\sqrt{\left(25-x^2\right)} \text {, then } \lim _{x \rightarrow 1}\left\{\frac{f(x)-f(1)}{x-1}\right\} \text { equals }}

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\mathrm{ \lim _{x \rightarrow 1}\left(\frac{x^2-1}{x \ln x}\right) \text { equals }}

Concepts Covered - 1

L’ Hospital’s Rule

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'(x)}{g'(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'(x)}{g'(x)}=\lim_{x\rightarrow a}\;\frac{f''(x)}{g''(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.

\\\text{(i)}\;\;\;\lim_{x\rightarrow 0}\;\frac{\sin x}{x}=\lim_{x\rightarrow 0}\;\frac{\cos x}{1}=1\\\\\text{(ii)}\;\;\;\lim_{x\rightarrow \infty}\;\frac{\log_e x}{x}=\lim_{x\rightarrow \infty}\;\frac{1/x}{1}=0\\\\\text{(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}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left ( \frac{\infty}{\infty}\;\text{form} \right )\\\\\text{\;}\;\;\;\;=\lim_{x\rightarrow \infty }\;\frac{1-\sin x}{1-\cos x},\;\text{which\;is\;cannot be\;calculated}\\\\\text{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

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L’ Hospital’s Rule

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