Question

$li{m}_{x\to 0}\frac{(e^{1/x}-1)}{(e^{1/x}+1)}$

• A. 0
• B. 1
• C. -1
• D. Does not exist

Answer 1

The correct option is D

Does not exist

f(x)=$\frac{(e^{1/x}-1)}{\left(e^{1/x}\right)+1}$, then

$li{m}_{x\to 0+}$ f(x) = $li{m}_{h\to 0}$$\frac{(e^{1/h}-1)}{(e^{1/h}+1)}$ =$li{m}_{h\to 0}$$\frac{(e^{1/h}-1)}{\left(e^{1/h}\right)+1}$ = $li{m}_{h\to 0}$$\frac{e^{1/h}\times (1-(\frac{1}{e^{1/h}})}{e^{\frac{1}{h}}\times (1+\frac{1}{e^{1/h}})}$ = 1
Similarly $li{m}_{x\to 0-}$f(x) = -1. Hence limit does not Exist