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Question: \(\lim_{x \rightarrow 0}\frac{a^{x} - 1}{\sqrt{1 + x} - 1}\) is equal to...

lim⁡x→0ax−11+x−1\lim_{x \rightarrow 0}\frac{a^{x} - 1}{\sqrt{1 + x} - 1} is equal to

A

2log⁡ea2\log_{e}a

B

12log⁡ea\frac{1}{2}\log_{e}a

C

alog⁡e2a\log_{e}2

D

None of these

Answer

2log⁡ea2\log_{e}a

Explanation

Solution

lim⁡x→0ax−11+x−1=lim⁡x→0ax−11+x−1.1+x+11+x+1\lim_{x \rightarrow 0}\frac{a^{x} - 1}{\sqrt{1 + x} - 1} = \lim_{x \rightarrow 0}\frac{a^{x} - 1}{\sqrt{1 + x} - 1}.\frac{\sqrt{1 + x} + 1}{\sqrt{1 + x} + 1}

= lim⁡x→0(ax−1)(1+x+1)1+x−1\lim_{x \rightarrow 0}\frac{(a^{x} - 1)\left( \sqrt{1 + x} + 1 \right)}{1 + x - 1} = lim⁡x→0(ax−1x).(1+x+1)\lim_{x \rightarrow 0}\left( \frac{a^{x} - 1}{x} \right).\left( \sqrt{1 + x} + 1 \right)

= (lim⁡x→0ax−1x).(lim⁡x→0(1+x+1))\left( \lim_{x \rightarrow 0}\frac{a^{x} - 1}{x} \right).\left( \lim_{x \rightarrow 0}\left( \sqrt{1 + x} + 1 \right) \right) = (log⁡ea).(2)(\log_{e}a).(2) =2log⁡ea2\log_{e}a.