Question

plz solve without using l hopital law

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Answer 1

GIVEN :–

• A limit –

$\\ \implies\bf \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ \sqrt{1 + x} - 1}$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let –

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ \sqrt{1 + x} - 1}$

• Now rationalization –

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ \sqrt{1 + x} - 1} \times \dfrac{ \sqrt{1 + x} + 1 }{\sqrt{1 + x} + 1}$

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ (\sqrt{1 + x} - 1)(\sqrt{1 + x}+1)} \times (\sqrt{1 + x}+1)$

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ (\sqrt{1 + x} )^{2} - (1)^{2}} \times (\sqrt{1 + x}+1)$

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ (1 + x - 1)} \times (\sqrt{1 + x}+1)$

$\\ \implies\bf P = \lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{x} \times (\sqrt{1 + x}+1)$

• Now Using L'HOSPITAL rule –

$\\ \implies\bf P = \lim_{x \to0} \bigg[ \dfrac{ \bigg(\dfrac{1}{2 + x} \bigg) \times(\sqrt{1 + x}+1) +[log(2 + x) - log(2)] \bigg[ \dfrac{1}{2 \sqrt{1 + x} } \bigg]}{1} \bigg]$

• Put limit –

$\\ \implies\bf P = \bigg[ \dfrac{ \bigg(\dfrac{1}{2 + 0} \bigg) \times(\sqrt{1 + 0}+1) +[log(2 + 0) - log(2)] \bigg[ \dfrac{1}{2 \sqrt{1 + 0} } \bigg]}{1} \bigg]$

$\\ \implies\bf P = \bigg[ \dfrac{ \bigg(\dfrac{1}{2} \bigg) \times(\sqrt{1}+1) +[log(2) - log(2)] \bigg[ \dfrac{1}{2 \sqrt{1}} \bigg]}{1} \bigg]$

$\\ \implies\bf P = \bigg[ \dfrac{ \bigg(\dfrac{1}{2} \bigg) \times(\sqrt{1}+1) +0}{1} \bigg]$

$\\ \implies\bf P = \bigg[ \dfrac{ \bigg(\dfrac{1}{2} \bigg) \times(1+1) }{1} \bigg]$

$\\ \implies\bf P = \bigg[ \dfrac{ \bigg(\dfrac{1}{2} \bigg) \times(2) }{1} \bigg]$

$\\ \implies\bf P = \bigg(\dfrac{1}{2} \bigg)\times (2)$

$\\ \implies\bf P =1$

• Hence –

$\\ \large \implies \red{ \boxed{\bf\lim_{x \to0} \dfrac{ log(2 + x) - log(2) }{ \sqrt{1 + x} - 1}=1}}$

Answer 2

Step-by-step explanation:

Given:

$lim \: x - > 0 \: \frac{ log(2 + x) - log2 }{ \sqrt{1 + x} - 1 }$

To find: Evaluate using L'Hospital rule

Solution:

• While applying L'HOSPITAL rule,Ni need to simply the function,just check for 0/0 form.
• The method is shown below:

By putting x=0

$\frac{ log(2 + 0) - log2 }{ \sqrt{1 + 0} - 1 } \\ \\ = > \frac{ log(2) - log(2) }{1 - 1} = > \frac{0}{0}$

Thus,apply L'HOSPITAL rule; Take derivative of numerator and denominator ,then apply limit

$lim \: x - > 0 \frac{ \frac{d}{dx} \bigg(log(2 + x) - log2 \bigg) } { \frac{d}{dx} \bigg(\sqrt{1 + x} - 1 \bigg) } \\ \\ lim \: x - > 0 \: \frac{ \frac{1}{2 + x} .1 - 0}{ \frac{1}{2 \sqrt{1 + x} } .1-0} \\ \\ simplify \\ \\ lim \: x - > 0 \: \frac{2 \sqrt{1 + x} }{2 + x} \\ \\ apply \: limit \\ \\ \: = > \frac{2 \sqrt{1 + 0} }{2 + 0} \\ \\ = > \frac{2 \sqrt{1} }{2} \\ \\ = > \frac{2}{2} \\ \\ = >1$

Thus,

$\bold{lim \: x - > 0 \: \frac{ log(2 + x) - log2 }{ \sqrt{1 + x} - 1 } = 1}$

Hope it helps you.

*Note: Don't do questions of limit using L'HOSPITAL rule in board exams.

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