Question

limit as x approaches zero is sin^2x/x... find the limit of the trigonometric function

Answer 1

Well, since plugging in 0 gets you the indeterminate form 0/0 you can use l'hopital's rule to find the limit

Answer 2

what is that

Answer 3

You haven't been taught L'Hopital's Rule yet?
Hmm darn, I guess we're suppose to use Squeeze Theorem or something else here.
Hmm.

Answer 4

Hmm yah i think that works :)

Answer 5

\[\Large \lim_{x\to0}\frac{\sin^2x}{x} \qquad=\qquad \lim_{x\to0}\frac{\sin x}{x}\cdot \lim_{x\to0}\sin x\]That first limit on the right is one you want to remember! :O

Answer 6

Welp, deleted that by mistake, what I said was change to (sinx/x) * sinx and solve (knowing that sinx/x = 1)

Answer 7

i get that sine of x over x is one but then one happens to the sine of x we took out

Answer 8

Plug in 0 for x and you get your limit :)

Answer 9

so would the answer be 3pi/2