Question

$ \displaystyle\lim_{x\rightarrow0} \frac{1}{3-2^{\frac{1}{x}}}$ is equal to

• A. 0
• B. $1$
• C. $\frac{1}{2}$
• D. $\frac{1}{3}$
• E. $-\infty$

Answer 1

Answer: (D)

$\displaystyle \lim _{x \rightarrow 0^{-}} \frac{1}{3-2^{1 / x}}$
Put $x =0-h $
$\therefore \displaystyle \lim _{h \rightarrow 0} \frac{1}{3-2^{\frac{1}{0-h}}}$
$ =\displaystyle \lim _{h \rightarrow 0} \frac{1}{3-2^{-\frac{1}{h}}} $
$=\frac{1}{3-2^{-\frac{1}{0}}}=\frac{1}{3-2^{-\infty}} $
$=\frac{1}{3-0}=\frac{1}{3}$