Limit of rational trig function

[prisma zealot]

does someone know how I would solve this limit. I multiplied the fraction by the conjugate 1 - cos and got theta^2(1+cos) over sin^2 but now I don't know what to do.

[silent basin]

Do u know L'hopitals?

Hmm actually

[prisma zealot]

im only in the first unit of calculus

[silent basin]

U just need to know

What $\lim_{\theta \to 0} \frac{\sin(\theta)}{\theta}$ is

[drowsy arrow]

992qqoloy

[prisma zealot]

i know what that is but how do I get sin on the top of the fraction

[silent basin]

U don't have to

use limit properties too I guess

Like how $\lim f(x) + g(x) = \lim f(x) + \lim g(x)$

[drowsy arrow]

992qqoloy

[silent basin]

That one won't help here

Just an example

[prisma zealot]

oh

I think I got Thank you. Just had to separate the sin^2 and divide it by theta squared.

[silent basin]

Yeah

[deep python]

$\frac{t^2}{1-cos(t)}
= \frac{t^2 (1+cos(t)}{(1-cos(t))(1+cos(t)}
= \frac{t^2(1+cos(t))}{1-cos^2(t)}
= (\frac{t}{sin(t)})^2 \times (1+cos(t))$

[drowsy arrow]

Daddy_314

[deep python]

Since the limit of sin(t)/t is equal to 1 at zero (by using the definition of the derivative)
By product of limits we get
1 times (1 + cos(0)) = 2

Of course there are many other ways to do it

[prisma zealot]

.solved

[crisp iron]

Solved

Post marked as solved by @prisma zealot.

Use .unsolved if this was a mistake.

[pseudo plover]

Isn't 1-cos(ø)/ø^2 = 1/2?

[deep python]

Yes, here we have the inverse of that