Arctan Limit of Cos(x) and e^x without L'Hopital's Rule

[genevievelily]

Homework Statement

lim x --> 0 (arctan(cosx))/(e^x)

a) infinity, b) pi/2, c) pi/4, pi/4 d) -infinity, e) -pi/2

Homework Equations

The Attempt at a Solution

You cannot use hospitals rule because its not indeterminate. e^x will approach 1, but I cannot figure out the numerator. Cos(x) will equal 1, but arctan of 1? Any help would be appreciated, exam coming up.

Thanks!

 

[vela]

Since $\tan x = \frac{\sin x}{\cos x}$, you're looking for the angle x such that
$$\frac{\sin x}{\cos x} = 1 \Rightarrow \sin x = \cos x.$$ This is something you need to know off the top of your head, and if you don't remember it, you look it up.

 

[genevievelily]

oh haha whoops definitely should have thought of that, thanks!