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Question: Let ƒ(x) = \(\left\{ \begin{matrix} \sin x, & x \neq n\pi \\ 2, & x = n\pi \end{matrix} \right.\ \),...

Let ƒ(x) = {sinx,xnπ2,x=nπ \left\{ \begin{matrix} \sin x, & x \neq n\pi \\ 2, & x = n\pi \end{matrix} \right.\ , where n ∈ I and

g (x) = {x2+1,x23,x=2 \left\{ \begin{matrix} x^{2} + 1, & x \neq 2 \\ 3, & x = 2 \end{matrix} \right.\ , then limx0\lim_{x \rightarrow 0}g [ƒ(x)] is –

A

1

B

0

C

3

D

Does not exist

Answer

1

Explanation

Solution

g [ƒ(x)] = {[f(x)]2+1,x23,x=2\left\{ \begin{array} { c c } { [ \mathrm { f } ( \mathrm { x } ) ] ^ { 2 } + 1 , } & \mathrm { x } \neq 2 \\ 3 , & \mathrm { x } = 2 \end{array} \right.

∴ g [ƒ (x)] = sin2 x + 1, x ≠ nπ

3, x = nπ

R.H.L. = limh0\lim _ { h \rightarrow 0 } g [ƒ(0 + h)] = limh0\lim _ { h \rightarrow 0 } (sin2 h + 1) = 1

L.H.L = limh0\lim _ { h \rightarrow 0 } g [ƒ(0 – h)] = limh0\lim _ { h \rightarrow 0 } (sin2 h + 1) = 1

limh0\lim _ { h \rightarrow 0 } g[ƒ (x)] = 1.