Question

Help with limts please

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin ^{2}x }{ x }\]

Answer 2

The sin thing throws me off. I remember that sin^2=1-cos^2 but i might be wrong

Answer 3

\[\frac{ \sin ^{2x} }{ x }=\frac{ 1-\cos ^{2}x }{ x }\]

Answer 4

Oh and the answer in the book is 0

Answer 5

\[\large \lim_{x \rightarrow 0}\frac{ \sin ^{2}x }{ x } = \lim_{x \rightarrow 0}\frac{ \sin x }{ x }*\sin x\]and you might know the limit of sin x/x...

Answer 6

Using limit laws\[\large \lim_{x \rightarrow 0}\frac{ \sin x }{ x }*\sin x = \lim_{x \rightarrow 0}\frac{ \sin x }{ x }*\lim_{x \rightarrow 0}\ \sin x\]

Answer 7

\[\frac{ \sin ^{2} x }{ x }=1\]
right?

Answer 8

\[\large \lim_{x \rightarrow 0}\frac{ \sin x }{ x }*\lim_{x \rightarrow 0}\ \sin x\]and you know\[\large \lim_{x \rightarrow 0}\frac{ \sin x }{ x } = 1\]so\[\large 1*\lim_{x \rightarrow 0}\ \sin x\]

Answer 9

1*0=0 :D

Answer 10

thank you so much for the explanation :D

Answer 11

Welcome.

Answer 12

:D ty to @IrishBoy123 too

Answer 13

^