Find the limit _ x 0 1 + x - 1 x . - Vidyadip

Question

Find the limit $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - 1}}{x}.$

✓

Answer

Put y = 1 + x, then y $\rightarrow$ 1 as x $\rightarrow$ 0.
$\therefore \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - 1}}{x} = \mathop {\lim }\limits_{y \to 1} \frac{{\sqrt y - 1}}{{y - 1}}$
$ = \mathop {\lim }\limits_{y \to 1} \frac{{{y^{\frac{1}{2}}} - {1^{\frac{1}{2}}}}}{{y - 1}}$
$ = \frac{1}{2}{(1)^{\frac{1}{2} - 1}}\left[ {\because \mathop {\lim }\limits_{x \to a} \frac{{{x^n} - {a^n}}}{{x - a}} = n{a^{n - 1}}} \right]$
$= \frac { 1 } { 2 }$

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