Question

This limit should be simple but I can't figure it out!
\[\lim_{t\rightarrow0}~~\frac{t^2}{\sin^2(t)}\]

Answer 1

What? \(\sin\) of what variable? Also the limit variable is \(n\)?

Answer 2

sorry i guess i wrote it kinda silly

Answer 3

\[\lim_{t\rightarrow0}~~\frac{t^2}{\sin^2(t)}\]

Answer 4

so...

Answer 5

I think the limit is 1. Use L'hopital rule and take the derivative of the top and the derivative of the bottom separately and then blog in 0 when you can't take the derivative anymore which them would equal to 2/2 which is 1.

Answer 6

i did use lhopital and you get
\[\frac{2t}{2\sin\cos}\]

Answer 7

and 2sincos is just 2sin(2\(\theta\))

Answer 8

and so im back to 0/0

Answer 9

*facepalm* no you are right. i should have done lhopitals twice. But i got 2/4

Answer 10

the denominator will be -sin(t)sin(t) + cos(t)cos(t) -sin(t)sin(t) + cos(t)cos(t) so when you blog in 0 it would equal to 0 + 1 - 0 + 1 so it would be 2.

Answer 11

\[
\frac{t^2}{\sin^2(t)} = \left(\frac{\sin t}{t}\right)^{-2}
\]Since we know the inner limit exists we can do this: \[
\lim_{t\to 0}\left(\frac{\sin t}{t}\right)^{-2}=\left(\lim_{t\to 0}\frac{\sin t}{t}\right)^{-2} = (1)^{-2} = 1
\]

Answer 12

are you sure you can do that?

Answer 13

ive never heard of this rule before

Answer 14

Really, you haven't?

Answer 15

What rules have you heard of?

Answer 16

It's just the division rule twice.

Answer 17

\[
\lim_{x\to a}f(x) = L \quad \text{and}\quad \lim_{x\to a}g(x) = K\implies \lim_{x\to a}\frac{f(x)}{g(x)} =\frac{L}{K}
\]First let \(f(x) = 1\), which means \(\lim_{x\to a}\frac{1}{g(x)} =\frac{1}{K}\)
Then let \(f(x) = 1/g(x)\), which means \(\lim_{x\to a}\frac{1}{{g(x)}^2} =\frac{\frac{1}{K}}{K} = \frac{1}{K^2}\)