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)((\cos(x)-1)/(x^2))=\lim _(x\to0)((-\sin(x))/(2x))=\lim _(x\to0)((-\cos(x))/(2))=(-\cos (0))/(2)=-(1)/(2)`

And for the second problem:

`\lim _(x\to0+)(((2x+1)^5)/(x))=\lim _(x\to0+)((10(2x+1)^4)/(1))=10(2(0)+1)^4=10`