Question

using l'hopital rule a) compute lim x-> 0 (cos(x)-1)/x^2b) compute lim x->0+ (2x+1)^5/x

Answer 1

We will solve as follows:

$\lim _{x\to0}(\frac{\cos(x)-1}{x^2})=\lim _{x\to0}(\frac{-\sin(x)}{2x})=\lim _{x\to0}(\frac{-\cos(x)}{2})=\frac{-\cos (0)}{2}=-\frac{1}{2}$

And for the second problem:

$\lim _{x\to0+}(\frac{(2x+1)^5}{x})=\lim _{x\to0+}(\frac{10(2x+1)^4}{1})=10(2(0)+1)^4=10$