Question

If the function defined by
$f(x) =
\begin{cases} \left(x^{2}+e^{\frac{1}{2-x}}\right)^{-1}, & \text{for} x \neq 2 \\ k, & \text{for x = 2} \end{cases}$
is right continuous at $x = 2$, then $k =$

• A. $-\frac{1}{4}$
• B. 0
• C. $\frac{1}{4}$
• D. 1

Answer 1

Answer: (C)

We have
$f(x) =
\begin{cases} \left(x^{2}+e^{\frac{1}{2-x}}\right)^{-1}, & \text{for} x \neq 2 \\ k, & \text{for x = 2} \end{cases}$
is right continuous at right of $x =2$
$\therefore \displaystyle\lim _{x \rightarrow 2^{+}}\left(x^{2}+e^{\frac{1}{2-x}}\right)^{-1}=k$
$\Rightarrow k= \displaystyle\lim _{h \rightarrow 0}\left[(2+h)^{2}+e^{\frac{1}{2-(2+h)}}\right]^{-1}$
$=\displaystyle\lim _{h \rightarrow 0}\left((2+h)^{2}+e^{-\frac{1}{h}}\right)^{-1}$
$=\displaystyle\lim _{h \rightarrow 0}\left((2+h)^{2}+\frac{1}{e^{\frac{1}{h}}}\right)^{-1}$
$=(4)^{-1}=\frac{1}{4}$