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

asked Dec 2, 2023 18.6k views

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6 votes

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:

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

Asaf Hananel answered Dec 6, 2023

3.7k points

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