JEE MainMathematicsEasyLimits, continuity and derivability

If $f(x) = \left\{ \begin{matrix}
3x - 1,\ \ \ \ x \geq 1 \\
2x + 3,\ \ \ \ x < 1
\end{matrix},g(x) = \right.\ \left\{ \begin{matrix}
3 - x &,x < 2 \\
2x - 3 &,x \geq 2
\end{matrix}, \right.\ \ then\ \lim_{x \rightarrow 2}\mspace{2mu} f(g(x)) =$

Show answer and solution

\ As\ x \rightarrow 2^{-} \Rightarrow 3 - x \rightarrow 1^{+}i.e.\ g(x) \rightarrow 1^{+}

\ As\ x \rightarrow 2^{+} \Rightarrow 2x - 3 \rightarrow 1^{+}i.e.\ g(x) \rightarrow 1^{+}

\therefore x \rightarrow 2 \Rightarrow g(x) \rightarrow 1^{+}

\lim_{x \rightarrow 2}\mspace{2mu} f\left( g(x) \right) = \lim_{g(x) \rightarrow 1^{+}}\mspace{2mu} f\left( g(x) \right)

= \lim_{x \rightarrow 1^{+}}\mspace{2mu} f(x)

= \lim_{x \rightarrow 1^{+}}\mspace{2mu}(3x - 1) = 2

Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx

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