Question

compute lim _ x➡️0 [ e^x -1 /root (1+x-1)]
answer my question correctly
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Answer 1

$\begin{gathered}\Large{\bold{\green{\underline{Formula \: Used \::}}}} \end{gathered}$

$\tt \:  ⟼ \: (1) \: \lim_{x\to0}\dfrac{ {e}^{x} - 1 }{x} = 1$

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$\large\underline\blue{\bold{Solution :- }}$

$\tt \:\lim_{x\to0}\dfrac{ {e}^{x} - 1 }{ \sqrt{1 + x} - 1 }$

☆ On substituting directly x = 0, we get indeterminant form

$\tt \:  = \tt \:\lim_{x\to0}\dfrac{ {e}^{x} - 1 }{ \sqrt{1 + x} - 1 } \times \dfrac{ \sqrt{1 + x} + 1 }{ \sqrt{1 + x} + 1 }$

$\tt \: = \: \lim_{x\to0}\dfrac{ ({e}^{x} - 1 ) \times ( \sqrt{1 + x} + 1) }{ {(\sqrt{1 + x})}^{2} - {1}^{2} }$

$\tt \: = \: \lim_{x\to0}\dfrac{ ({e}^{x} - 1 ) }{ \cancel1 + x - \cancel1 } \times \lim_{x\to0}( \sqrt{1 + x} + 1)$

$\tt \:  = \: 1 \times ( \sqrt{1 + 0} + 1)$

$\tt \:  = \: 2$

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