The function defined by $f(x) = \left\{ \begin{array}{r} \left( x^{2} + e^{\frac{1}{2 - x}} ight)^{- 1}\ x
eq 2 \ k,\ \ \ \ \ \ \ \ \ \ x = 2 \end{array} ight.\$ is continuous from right at the point $x = 2$, then $k$ is equal to

Question

The function defined by $f(x) = \left\{ \begin{array}{r} \left( x^{2} + e^{\frac{1}{2 - x}} ight)^{- 1}\ x 
eq 2 \ k,\ \ \ \ \ \ \ \ \ \ x = 2 \end{array} ight.\$ is continuous from right at the point $x = 2$, then $k$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Since, $\lim_{x{\rightarrow 2}^{+}}{f(x) = f(2) = k}$
$\Rightarrow \ \ \ k = \lim_{h \rightarrow 0}{f(2 + h)}$
$\Rightarrow \ \ \ k = \underset{h \rightarrow 0}{lim}\left\lbrack (2 + h)^{2} + e^{\frac{1}{2 - (2 + h)}} \right\rbrack^{- 1}\$
$\Rightarrow \ \ \ \underset{h \rightarrow 0}{lim}\ \ \ \left\lbrack 4 + h^{2} + 4h + e^{\frac{- 1}{h}} \right\rbrack^{- 1} = \frac{1}{4}$

Q. The function defined by $f(x) = \left\{ \begin{array}{r} \left( x^{2} + e^{\frac{1}{2 - x}} \right)^{- 1}\ x \neq 2 \\ k,\ \ \ \ \ \ \ \ \ \ x = 2 \end{array} \right.\$ is continuous from right at the point $x = 2$, then $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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