Module calculus problem

Question

Find 

Options:

1. 0/0
2. 1
3. 0
4. 1/2

Answer 1

Whenever you see a limit question, the first thing you should always do is try the value in the expression.

0³ + sin 0 0 0

-------------- = ----- = -----

cos 0 - (0 - 1)² 1 - 1 0

But we all know that this won't do.

This limit question seems to be a job for L'Hopital's Rule. Before applying L'Hopital's Rule, you must always confirm that it meets the conditions:

1. Both the numerator and the denominator must equal 0 when the limit value is substituted.
2. Both the numerator and the denominator must be differentiable at the limit value.

Our expression meets both conditions. So we differentiate the numerator and denominator separately and try again.

x³ + sin x becomes 3x² + cos x

cos x - (x - 1)² becomes -sin x - 2(x - 1)

We try to substitute 0 again---

Numerator = 3 • 0² + cos 0 = 0 + 1 = 1

Denominator = -sin 0 - 2(0 - 1) = 0 - 2(-1) = 2

Put it together, and the limit equals ¹/₂.