Question

The right hand and left hand limit of the function are respectively

• A. 1 and 1
• B. 1 and -1
• C. -1 and -1
• D. -1 and 1

Answer 1

Answer: (B)

1 and -1

Answer 2

$LHL =\displaystyle\lim _{x \rightarrow 0^{-}}\left(\frac{e^{1 / x}-1}{e^{1 / x}+1}\right)$
$=\frac{e^{-\infty}-1}{e^{-\infty}+1}=\frac{0-1}{0+1}=-1$
Dividing both numerator and denominator by $e^{1 / x}$, we get
$\displaystyle\lim _{x \rightarrow 0}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$RHL =\displaystyle\lim _{x \rightarrow 0^{+}}\left(\frac{1-e^{-1 / x}}{1+e^{-1 / x}}\right)$
$=\frac{1-e^{-\infty}}{1+e^{-\infty}}=\frac{1-0}{1+0}=1$