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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?

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  1. amistre64
    • one year ago
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    where are you stuck at?

  2. hpfan101
    • one year ago
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    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
    • one year ago
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    I think the fact that it has a sine in the function is what making me confused.

  4. hpfan101
    • one year ago
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    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
    • one year ago
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    what is your definition of the sgn function?

  6. amistre64
    • one year ago
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    |dw:1442784418003:dw|

  7. amistre64
    • one year ago
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    |dw:1442784453858:dw|

  8. amistre64
    • one year ago
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    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
    • one year ago
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    so this question reduces to finding out if the value is approaching -1 or 1 from the given direction and value.

  10. hpfan101
    • one year ago
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    The definition of sgn based on the textbook was "signum" (or sign) function.

  11. amistre64
    • one year ago
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    right, and its value is either -1, 0, or 1 depending on the sign of its argument

  12. amistre64
    • one year ago
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    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
    • one year ago
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    Oh, I see. Alright, thank you.

  14. amistre64
    • one year ago
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    youre welcome

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