Answer:
[tex]\lim_{x\rightarrow 0}g(x)[/tex] does not exist.
D is correct.
Step-by-step explanation:
We are given,
[tex]g(x)=\left\{\begin{matrix} \sin\left ( \dfrac{\pi}{x} \right )& \ \ \ if\ \ x\neq 0\\ 1& \ \ \ if\ \ x\neq 0\end{matrix}\right.[/tex]
For continuous function, LHL=RHL=f(0)
Using squeeze theorem,
If [tex]g(x)\leq f(x)\leq h(x)[/tex]
then [tex]\lim_{x\rightarrow a}g(x)\leq \lim_{x\rightarrow a}f(x)\leq \lim_{x\rightarrow a}h(x)[/tex]
As we know
[tex]-1\leq \sin\left ( \dfrac{\pi}{x} \right )\leq 1[/tex]
Apply Limit both sides
[tex]\lim_{x\rightarrow 0}(-1)\leq \lim_{x\rightarrow 0}\sin\left ( \dfrac{\pi}{x} \right )\leq \lim_{x\rightarrow a}1[/tex]
[tex]\lim_{x\rightarrow 0}(-1)=-1[/tex]
[tex]\lim_{x\rightarrow 0}(1)=1[/tex]
[tex]-1\neq 1[/tex]
[tex]LHL\neq RHL=g(0)[/tex]
If [tex]LHL\neq RHL[/tex] then limit does not exist.
Hence, [tex]\lim_{x\rightarrow 0}g(x)[/tex] does not exist.
D is correct.
