## hpfan101 one year ago 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?

1. amistre64

where are you stuck at?

2. hpfan101

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?

3. hpfan101

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

4. hpfan101

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

5. amistre64

what is your definition of the sgn function?

6. amistre64

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7. amistre64

|dw:1442784453858:dw|

8. amistre64

sgn(u) = -1 if u < 0 sgn(u) = 0 if u = 0 sgn(u) = 1 if u > 0 sin(x) alternates between -1 and 1 so its composite here will do the same ...

9. amistre64

so this question reduces to finding out if the value is approaching -1 or 1 from the given direction and value.

10. hpfan101

The definition of sgn based on the textbook was "signum" (or sign) function.

11. amistre64

right, and its value is either -1, 0, or 1 depending on the sign of its argument

12. amistre64

sgn(-32.484843) = -1, since the argument is negative sgn(29.1823) = 1, since the argument is positive sgn(0) = 0 since zero is neither positive or negative.

13. hpfan101

Oh, I see. Alright, thank you.

14. amistre64

youre welcome