Expert: Steve Holleran - 4/7/2007

Question
limit as x goes to zero of (e^x-cos(x)-2x)/(x^2-2x)
and how to solve it

Answer
Hi Craig,

Okay, this looks like a candidate for using L'Hopitals Rule.  This first thing you ever do when evaluating a limit is to substitute the value x approaches, and check the result.

So, if we do that here, we have:

lim, x-->0, (e^x - cos x - 2x) / (x^2 - 2x)

=           (e^0 - cos 0 -2*0) / (0^2 - 2*0)

=            (1 - 1 - 0) / (0 - 0) = 0 / 0 .

So now we must use L'Hopitals Rule, which says that if you get this form, take the derivatives separately of the top and bottom of the fraction, then take the limit again:

lim, x-->0, (e^x - cos x - 2x) / (x^2 - 2x)

= lim, x-->0, (e^x - (-sin x) - 2) / (2x - 2)

=             (e^0 + sin 0 - 2) / (0 - 2)

=             ( 1  +   0   - 2) / (-2)

=                      -1 / -2 = 1/2.

(Just a thought here, Craig--some students make the mistake of using the Quotient Rule to take the derivative here.  All you want to do is verify that you have an eligible form for LH's Rule--0/0 or inf/inf--and then take the top derivative and bottom derivative separately.  After that, re-submit the x-value.  If you get 0/0 or inf/inf again, just repeat the process until you get an answer).

I hope this helps you out.

Steve Holleran