_ x 0 e ^ 1 / x -1 e ^ 1 / x +1 = - Vidyadip

MCQ

$\lim _{x \rightarrow 0} \frac{ e ^{1 / x}-1}{ e ^{1 / x}+1}=$

A
$0$
B
1
C
-1
✓
Does not exist

✓

Answer

Correct option: D.

Does not exist

(D)
Let $f(x)=\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$, then
$\lim _{x \rightarrow 0^{+}} f^{\prime}(x)=\lim _{h \rightarrow 0}\left(\frac{e^{1 / h}-1}{e^{1 / h}+1}\right)=\lim _{h \rightarrow 0} \frac{e^{1 / h}\left(1-\frac{1}{e^{1 / h}}\right)}{e^{1 / h}\left(1+\frac{1}{e^{1 / h}}\right)}=1$
Similarly, $\lim _{x \rightarrow 0^{-}} f (x)=-1$.
Hence, limit does not exist.

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