Question

4) lim X=0 2^x-1/✓1+x-1
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Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ {2}^{x} - 1 }{ \sqrt{1 + x} - 1}}$

If we substitute directly x = 0, we get

$\rm \: = \: \: \dfrac{ {2}^{0} - 1 }{ \sqrt{1 + 0} - 1}$

$\rm \: = \: \: \dfrac{1 - 1}{1 - 1}$

$\rm \: = \: \: \dfrac{0}{0} \: which \: is \: meaningless$

So,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ {2}^{x} - 1 }{ \sqrt{1 + x} - 1}}$

On rationalizing the denominator, we get

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \dfrac{ {2}^{x} - 1 }{ \sqrt{1 + x} - 1} \times \dfrac{ \sqrt{1 + x} + 1}{ \sqrt{1 + x} + 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \dfrac{( {2}^{x} - 1 )( \sqrt{1 + x} + 1)}{ (\sqrt{1 + x}) {}^{2} - {1}^{2} }$

$\red{\bigg \{ \because \:(x + y)(x - y) = {x}^{2} - {y}^{2} \bigg \}}$

$\rm \: = \: \: \displaystyle\lim_{x \to 0} \dfrac{( {2}^{x} - 1)( \sqrt{1 + x} + 1)}{1 + x - 1}$

$\rm \: = \: \: \displaystyle\lim_{x \to 0} ( \sqrt{1 + x} + 1) \times \displaystyle\lim_{x \to 0} \frac{ {2}^{x} - 1}{x}$

We know,

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \frac{ {a}^{x} - 1}{x} = loga}}$

$\rm \: = \: \: (1 + 1) \times log(2)$

$\rm \: = \: \: 2 \: log2$

Hence,

$\red{\rm :\longmapsto\:\displaystyle\lim_{x \to 0} \frac{ {2}^{x} - 1 }{ \sqrt{1 + x} - 1} = 2 \: log2}$

Additional Information :-

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \frac{tanx}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \frac{log(1 + x)}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to 0} \frac{ {e}^{x} - 1}{x} = 1}}$

$\boxed{ \rm{ \displaystyle\lim_{x \to a} \frac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{n - 1}}}$