Question
Let \[f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,{5^{\frac{1}{x}}},\,\,\,\,\,\,\,x < 0\\
\lambda \left[ x \right],\,\,\,\,x \ge 0
\end{array} \right.{\rm{ \,and\,\, }}\lambda \, \in \,R\] then at $$x = 0$$
A.
$$f$$ is discontinuous
B.
$$f$$ is continuous only, $$\lambda = 0$$
C.
$$f$$ is continuous only, whatever $$\lambda $$ may be
D.
none of these
Answer :
$$f$$ is discontinuous
Solution :
As we know,
A function $$f\left( x \right)$$ is said to be continuous at a point $$x = a$$ iff
$$\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right),$$ otherwise not continuous.
Thus $$f\left( x \right)$$ is continuous at $$x = a$$ iff
$$\mathop {\lim }\limits_{x \to {a^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} f\left( x \right) = f\left( a \right)$$
\[{\rm{Since, }}f\left( x \right) = \left\{ \begin{array}{l}
\,\,\,{5^{\frac{1}{x}}},\,\,\,\,\,\,\,x < 0\\
\lambda \left[ x \right],\,\,\,\,x \ge 0
\end{array} \right.{\rm{ \,and\,\, }}\lambda \, \in \,R\]
$$\eqalign{
& {\text{R}}{\text{.H}}{\text{.L}}{\text{. at }}x = 0:\mathop {\lim }\limits_{x \to {0^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \lambda \left[ x \right] = \mathop {\lim }\limits_{h \to 0} \lambda \left[ h \right] = 0 \cr
& {\text{L}}{\text{.H}}{\text{.L}}{\text{. at }}x = 0:\mathop {\lim }\limits_{x \to {0^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {0^ - }} {5^{\frac{1}{x}}} = \mathop {\lim }\limits_{h \to 0} {5^{ - \frac{1}{h}}} = {5^\infty } = \infty \cr
& {\text{and }}f\left( 0 \right) = \lambda \left[ 0 \right]0 \cr
& {\text{Since, L}}{\text{.H}}{\text{.L}}{\text{.}} \ne {\text{R}}{\text{.H}}{\text{.L}}{\text{.}} \cr} $$
$$\therefore \,f\left( x \right)$$ is not continuous.