lim as x->0 of e^sin(x)
I know how to solve similar limits involving trig functions and all, but this one doesn't make sense. If the limit as x->0 of e^x by itself is 0 and the limit of sin is 0 as well, then wouldn't the answer be 0? I know the answer to the above problem is 1, but why? How do you split up sin and e to solve the limit?