Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ x
eq n\pi} \ 2,\ \ \ \ x = n\pi \end{array} ight.\$, where $n \in I$ and $g(x) = \left\{ \begin{array}{r} x^{2} + 1,\ \ \ x
eq 2 \ 3,\ \ \ \ \ \ \ \ x = 2 \end{array} ight.\$, then $\lim_{x ightarrow 0}{g\lbrack f(x) brack}$ is

Question

Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ x 
eq n\pi} \ 2,\ \ \ \ x = n\pi \end{array} ight.\$, where $n \in I$ and $g(x) = \left\{ \begin{array}{r} x^{2} + 1,\ \ \ x 
eq 2 \ 3,\ \ \ \ \ \ \ \ x = 2 \end{array} ight.\$, then $\lim_{x ightarrow 0}{g\lbrack f(x)brack}$ is

Dr. Abubacker · Accepted Answer

(a)
$g\lbrack f(x)brack = \left\{ \begin{array}{r}
{\lbrack f(x)brack}^{2} + 1,\ \ \ f(x) 
eq 2 \
3,\ \ \ \ \ \ \ \ f(x) = 2
\end{array} ight.\$
$\Rightarrow \ \ \ \ g\lbrack f(x)brack = \ \left\{ \begin{array}{r}
\sin^{2}x + 1,\ \ \ x 
eq n\pi \
3,\ \ \ \ \ \ \ \ x = n\pi
\end{array} ight.\$
RHL$= \lim_{h ightarrow 0}{g\lbrack f(0 + h)brack}$
$= \lim_{h ightarrow 0}{\left( \sin^{2}{h + 1} ight) = 1}$
And LHL$= \lim_{h ightarrow 0}{g\lbrack f(0 - h)brack}$
$= \lim_{h ightarrow 0}{\left( \sin^{2}{h + 1} ight) = 1}$
$	herefore\ \ \ \ \ \ \ \ \lim_{x ightarrow 0}{g\lbrack f(x)brack} = 1\$

Q. Let $f(x) = \left\{ \begin{array}{r} \sin{x,\ \ x \neq n\pi} \\ 2,\ \ \ \ x = n\pi \end{array} \right.\$, where $n \in I$ and $g(x) = \left\{ \begin{array}{r} x^{2} + 1,\ \ \ x \neq 2 \\ 3,\ \ \ \ \ \ \ \ x = 2 \end{array} \right.\$, then $\lim_{x \rightarrow 0}{g\lbrack f(x)\rbrack}$ is

Explanation

Reviewed by Dr. Abubacker

Want more practice?