Let $$g(x)=sgn(\sin x)$$

a) Find each of the following limits or explain why it does not exist.

i. $$\lim_{x \rightarrow 0^+} g(x)$$
ii. $$\lim_{x \rightarrow 0^-} g(x)$$
iii. $$\lim_{x \rightarrow 0} g(x)$$
iv. $$\lim_{x \rightarrow \pi^-} g(x)$$
v. $$\lim_{x \rightarrow \pi^+} g(x)$$
$$\lim_{x \rightarrow \pi} g(x)$$

b) For which values of a does lim x approaches a for the function, g(x), not exist?

Question

Let $$g(x)=sgn(\sin x)$$

a) Find each of the following limits or explain why it does not exist. 

i. $$\lim_{x \rightarrow 0^+} g(x)$$
ii. $$\lim_{x \rightarrow 0^-} g(x)$$
iii. $$\lim_{x \rightarrow 0} g(x)$$
iv. $$\lim_{x \rightarrow \pi^-} g(x)$$
v. $$\lim_{x \rightarrow \pi^+} g(x)$$
$$\lim_{x \rightarrow \pi} g(x)$$

b) For which values of a does lim x approaches a for the function, g(x), not exist?

amistre64 · Answer

where are you stuck at?

hpfan101 · Answer

Well, I'm confused about the one-sided limits. When I substitute 0 into the equation, the sine of 0 is 0. So I'm not sure if I'm doing it right. Because wouldn't it be the same as x approaches 0 from the right or left?

hpfan101 · Answer

I think the fact that it has a sine in the function is what making me confused.

hpfan101 · Answer

Would we just substitute a value less than 0 to find the limit as x approaches 0 from the left. Like -0.01?

amistre64 · Answer

what is your definition of the sgn function?

amistre64 · Answer

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