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Find the limit \lim _{x \rightarrow 0} \frac{2^{x}-1}{(1+x)^{1 / 2}-1}=

Option: 1

{\log 2}


Option: 2

{\log 4}


Option: 3

{\log \sqrt{2}}


Option: 4

None of these


Answers (1)

 

 

Limit Using Expansion (Part 1) -

Limit Using Expansion (Part 1)

 

Using the expansion of some function is one of the easier methods to finding the limit of expression. The following expansion formulas which is also known as Taylor series, are very useful in evaluating various limits.

 
 

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\text{As we know }\\ a^{x}=1+x\left(\log _{e} a\right)+\frac{x^{2}}{2 !}\left(\log _{e} a\right)^{2}+\frac{x^{3}}{3 !}\left(\log _{e} a\right)^{3}+\ldots \ldots \ldots \\ (1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\ldots \cdots \cdots \\ \lim _{x \rightarrow 0} \frac{2^{x}-1}{(1+x)^{1 / 2}-1}=\frac{x \log_e2+\frac{x^{2}}{2 !}\left(\log _{e} 2\right)^{2}+\frac{x^{3}}{3 !}\left(\log _{e} 2\right)^{3}+\ldots \ldots \ldots }{\frac{1}{2} x+\frac{\frac{1}{2}(\frac{1}{2}-1)}{2 !} x^{2}+\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)}{3 !} x^{3}+\ldots \cdots \cdots}\\ =2 \log_e2\\ =\log_e4

Posted by

Kshitij

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