Hi i’m not sure how to do question 20 if u could explain how to do it that’d b great

  A)  -2

Step-by-step explanation:

The form is indeterminate at x=0, so L'Hopital's rule applies. The resulting form is also indeterminate at x=0, so a second application is required.

Let f(x) = x·sin(x); g(x) = cos(x) -1

Then f'(x) = sin(x) +x·cos(x), and g'(x) = -sin(x).

We still have f'(0)/g'(0) = 0/0 . . . . . indeterminate.

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Differentiating numerator and denominator a second time gives ...

  f''(x) = 2cos(x) -sin(x)

  g''(x) = -cos(x)

Then f''(0)/g''(0) = 2/-1 = -2

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I like to start by graphing the expression to see if that is informative as to what the limit should be. The graph suggests the limit is -2, as we found.

hi there!

the answer will be d: 2