BC Calculus Question

Test Preparation AP Test Preparation
1

<p>How do you do this problem: (non calc)</p>

<p>lim x->0 [e^x - cosx - 2x] / [x^2 - 2x]</p>

<p>the answer is 1/2</p>

<p>fyi, the problem is from 2003 BC exam</p>

2

<p>You have to use L`Hopital's Rule
1. Lim x-->0 e^x-cos x-2x/(x^2-2x)
2. Differentiate the top and bottom
3. Lim x-->0 e^x+sinx-2/(2x-2)
4. Sub in the 0 and 1/2 is the answer</p>