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  • Evaluate   <img alt="\mathrm{\lim _{x \rightarrow 1} \frac{1+e^{\frac{1}{x-1}}}{1-e^{\frac{1}{x-1}}}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim%20_%7Bx%20%5

Evaluate   \mathrm{\lim _{x \rightarrow 1} \frac{1+e^{\frac{1}{x-1}}}{1-e^{\frac{1}{x-1}}}}

Option: 1

0


Option: 2

-1


Option: 3

does not exist 


Option: 4

1


Answers (1)

best_answer

Let  \mathrm{t=\frac{1}{x-1} \rightarrow \infty}, Then 

\mathrm{\frac{1+e^t}{1-e^t}=\frac{e^{-t}+1}{e^t-1} \rightarrow \frac{1}{-1}=-1}

This is true assuming you mean \mathrm{x\rightarrow 1^{+}}.

If \mathrm{x\rightarrow 1^{-}}, then \mathrm{t\rightarrow -\infty} and the limit evaluates to 1.

Hence, the limit doesn't exist.

Posted by

chirag

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