Question

Explain why g(x)= {sin π/x if =0, 1 if x=0 , is not continuous a x=0

Answer 1

Answer:

$\lim_{x\rightarrow 0}g(x)$ does not exist.

D is correct.

Step-by-step explanation:

We are given,

$g(x)=\left\{\begin{matrix} \sin\left ( \dfrac{\pi}{x} \right )& \ \ \ if\ \ x\neq 0\\ 1& \ \ \ if\ \ x\neq 0\end{matrix}\right.$

For continuous function, LHL=RHL=f(0)

Using squeeze theorem,

If $g(x)\leq f(x)\leq h(x)$

then $\lim_{x\rightarrow a}g(x)\leq \lim_{x\rightarrow a}f(x)\leq \lim_{x\rightarrow a}h(x)$

As we know

$-1\leq \sin\left ( \dfrac{\pi}{x} \right )\leq 1$

Apply Limit both sides

$\lim_{x\rightarrow 0}(-1)\leq \lim_{x\rightarrow 0}\sin\left ( \dfrac{\pi}{x} \right )\leq \lim_{x\rightarrow a}1$

$\lim_{x\rightarrow 0}(-1)=-1$

$\lim_{x\rightarrow 0}(1)=1$

$-1\neq 1$

$LHL\neq RHL=g(0)$

If $LHL\neq RHL$ then limit does not exist.

Hence, $\lim_{x\rightarrow 0}g(x)$ does not exist.

D is correct.