Question

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

Answer 1

D

Step-by-step explanation: EDGE 2020

Answer 2

`\lim_(x\rightarrow 0)g(x)` does not exist.

D is correct.

Explanation:

We are given,

`g(x)=\left\{\begin{matrix} \sin\left ( (\pi)/(x) \right )& \ \ \ if\ \ x\\eq 0\\ 1& \ \ \ if\ \ x\\eq 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 ( (\pi)/(x) \right )\leq 1`

Apply Limit both sides

`\lim_(x\rightarrow 0)(-1)\leq \lim_(x\rightarrow 0)\sin\left ( (\pi)/(x) \right )\leq \lim_(x\rightarrow a)1`

`\lim_(x\rightarrow 0)(-1)=-1`

`\lim_(x\rightarrow 0)(1)=1`

`-1\\eq 1`

`LHL\\eq RHL=g(0)`

If
`LHL\\eq RHL` then limit does not exist.

Hence,
`\lim_(x\rightarrow 0)g(x)` does not exist.

D is correct.