Question

lim x approaches zero cosx^2-1/x

Answer 1

which is -sin(x) / x, which is -1

Answer 2

this is
\[\frac{cos^2(x)-1}{x}=\frac{-\sin^2(x)}{x}\]

Answer 3

since lim x-->0 of sinx / x is 1

Answer 4

oh no not quite

Answer 5

wow im off today my bad i was rushing!

Answer 6

hopitals rule

Answer 7

\[\lim_{x \rightarrow 0}\frac{\cos^2(x)-1}{x}=\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x} \cdot \lim_{x \rightarrow 0}(\cos(x)+1)=0 \cdot (\cos(0)+1)=0(1+1)=0(2)=0\]

Answer 8

-2sin(x) / 1 so lim is zero

Answer 9

same as
\[-\frac{\sin(x)}{x}\times \sin(x)\] right? first limit is 1second is 0 get 0

Answer 10

satellite isnt first limit -1... im pretty sure im right for that

Answer 11

no need for l'hopital here. limit is straight up 0

Answer 12

yeah it is -1 you are right. product is 0 in any case for sure

Answer 13

yup

Answer 14

well u can do either product of lims or hopital...wtvr u want

Answer 15

thanks