If $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\sin x,x \ne n{\rm{ }}\pi ,n \in Z}\\ {0,{\rm{other}}\,\,\,{\rm{wise}}} \end{array}} \right.$ and $g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {{x^2} + 1,x \ne 0,2}\\ {4,x = 0}\\ {5,x = 2} \end{array}} \right.$, then $\mathop {\lim }\limits_{x \to 0} g\left( {f\left( x \right)} \right)$ is

JEE MATHEMATICS MCQ: LIMITS

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f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {\sin x,x \ne n\mathrm{ }\pi ,n \in Z}\\ {0,\mathrm{other}\,\,\,\mathrm{wise}} \end{array}} \right.

and

g\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {{x^2} + 1,x \ne 0,2}\\ {4,x = 0}\\ {5,x = 2} \end{array}} \right.

, then

\lim\limits_{x \to 0} g\left( {f\left( x \right)} \right)

View answer and explanation

Correct option: A

Option A (1) is correct because as x approaches 0, sin(x) also approaches 0, and g(0) = 4. However, since x != n*pi, we use the other definition of g(x) which is x² + 1. Thus, the limit is 0² + 1 = 1.

JEE MATHEMATICS MCQ: LIMITS

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