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padilas [110]
3 years ago
6

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

Mathematics
1 answer:
stepladder [879]3 years ago
7 0

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.

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