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Let $ f(x) = \frac {x}{\sqrt{1 - \cos 2x}} $

(a) Graph $ f. $ What type of discontinuity does it appear to have at $ 0? $

(b) Calculate the left and right limits of $ f $ at $ 0. $ Do these values confirm your answer to part (a)?

a) Jump discontinuity

b) See explanation for solution.

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yet square. So when you read here, so we have f of X is equal to X over square it of one minus co sign two x When we draw this it looked something like this and you see that there is a jump discontinuity at zero part B. We're gonna be focusing on the bottom part. We're gonna use the trick identity for a double angle co sign only, get one minus co sign of two X is equal to one minus one minus two. Signed square, which is equal to two signs square. When we calculate the dirt, the limit as X goes to the positive side of zero for X over square of one minus co. Sign two acts. This becomes equal to the limit as X approaches zero positive for X over square root of to sign square root of to sign Sign of X, which is equal to one over square root of two limit as X approaches. Zero positive of X. Over sign of X. This is equal to one over square root of two. This is from the right side, so it's gonna be positive. And when we find the derivative the limit. Excuse me. Um, since it's gonna be from the left side, it's gonna be negative.