Question

Calculus Question:
lim (tan^2 x)/ x
x approaches 0

Answer 1

L'hopital rule

Answer 2

? i haven't learned that.. the only 2 identities they gave me were
sinx / x = 0 and (1-cosx)/x =1

Answer 3

nope, lim sinx/x when x approaches 0 is 1 , not 0

Answer 4

oh oops i swtiched the two. I should have written sinx / x =1 and (1-cosx)/x =0

Answer 5

ok, let see. long time no work without l'hopital. Let try once. what do you get so far?

Answer 6

well i have ((sinx/cosx) times (sinx/cosx)) / x
i dont think its right

Answer 7

\[\frac{tan^2}{x} =\frac{sin^2}{xcos^2}= \frac{sin}{x}\frac{sin}{cos^2}\] so, limit of it when x approaches to 0 is 0 since cos (0) =1 and sin(0) =0,

Answer 8

the question says there is an x after the tan^2. so it look like this \[(\tan^2x) / x\]

Answer 9

I just didn't type it out ( x), it 's still there, friend. I am sorry, I think you understand.

Answer 10

Another way to compute the limit: \(\tan x\approx x\) for \(x\) near 0, so you have
\[\lim_{x\to0}\frac{\tan^2x}{x}=\lim_{x\to0}\frac{x^2}{x}=\lim_{x\to0}x=0\]