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L’ Hospital’s Rule - Practice Questions & MCQ

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

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

  • 189 Questions around this concept.

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QuestionMEDIUM

If f(1)=1,f'(1)=2\; then\;      is

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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 :

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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)}:

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\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 }}

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\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}

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

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\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 }}

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\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 }}

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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

Study other Related Concepts

L’ Hospital’s Rule Current Topic

Limit
Left-Hand Limits and Right-Hand Limits
Algebra of Limits
Limit of Indeterminate Form and Algebraic limit
Limit of Algebraic function
Limit Using Expansion
Sandwich Theorem
Limits of Trigonometric Functions
Exponential and Logarithmic Limits
Limits of the form (1 power infinity)

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L’ Hospital’s Rule

Mathematics for Joint Entrance Examination JEE (Advanced) : Calculus

Page No. : 2.27

Line : 26

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