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  • <img alt="\lim_{z\rightarrow 0}\mathrm{\left ( \frac{1}{x}-\frac{1}{e^{x}-1} \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Clim_%7Bz%5Crightarrow%200%7D%5Cmathrm%7B%5Cleft%20%28

\lim_{z\rightarrow 0}\mathrm{\left ( \frac{1}{x}-\frac{1}{e^{x}-1} \right )}

Option: 1

\frac{1}{2}


Option: 2

1


Option: 3

2


Option: 4

0


Answers (1)

best_answer

Let us rewrite the expression in the limit as follows
\mathrm{ \frac{1}{x}-\frac{1}{e^x-1}=\frac{e^x-1-x}{x e^x-x} }
Using the Taylor expansion of \mathrm{ e^x }, which we recall is
\mathrm{ e^x=\sum_{i=0}^{\infty} \frac{x^i}{i !}, }
we further rewrite the numerator and denominator as

\mathrm{ e^x-1-x=\frac{x^2}{2}+\mathcal{O}\left(x^3\right)=x^2\left(\frac{1}{2}+\mathcal{O}(x)\right) }

\mathrm{ x e^x-x=x^2+\mathcal{O}\left(x^3\right)=x^2(1+\mathcal{O}(x)) }

where \mathrm{ \mathcal{O}\left(x^3\right) } denotes the terms of order higher then 2 in the variable \mathrm{ x }. Since \mathrm{ \lim _{x \rightarrow 0} \mathcal{O}(x)=0 } we can compute, using the theorem for product of limits, in the following manner

\mathrm{ \lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{e^x-1}\right)=\lim _{x \rightarrow 0}\left(\frac{e^x-1-x}{x e^x-x}\right)=\lim _{x \rightarrow 0} \frac{x^2\left(\frac{1}{2}+\mathcal{O}(x)\right)}{x^2(1+\mathcal{O}(x))}=\lim _{x \rightarrow 0} \frac{\frac{1}{2}+\mathcal{O}(x)}{1+\mathcal{O}(x)}=\frac{1}{2} }
Hence option 1 is correct.

Posted by

Gaurav

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