Question

Can anyone help me solve this question? I'm not sure how to handle it with the 5 and 8 shoved in there. Evaluate the limit as theta approaches zero of (sin5theta)/(theta +(tan8theta))

Answer 1

Have you learned l'hoptial's rule yet?

Answer 2

Yes I have, working with that, I thought the answer should be 1/2 and I got marked wrong.

Answer 3

It's not 1/2 (: Forgive me for any incorrect usage of definitions...I get them confused sometimes. But here goes a walk through;

Okay, so you know when a limit converges to 0 on top and bottom, that you should apply l'hoptial's rule. Step one. Apply l'hoptials rule. I'm going to see if this is where you're making your mistake.

What is the derivative of cos(5x)?

Answer 4

sin5theta/theta when theta approaches 0 is 5 by L'hospital rule... derivate if sin5x is 5*cos5x so answer is 5

Answer 5

The answer isn't five, either. :| Don't make it worse.

Answer 6

oh sory.. its 5/(1+8)

Answer 7

The derivative of cos(5x) is -5sin(5x), correct?

Answer 8

yup.. but u dont need to do dat here

Answer 9

Right, just making sure you understood trig derivatives.

When you apply l'hoptial's rule, your answer won't be zero when you check again. You should have 5 cos(5x)/(1+sec^2(8x)). Which, when you plug zero in, will end up being (5cos(0))/(1+8sec^2(0)).

Rewrote the paragraph; made one too many typos. Haha. So the answer is basic trig from there.

Answer 10

I am trying to understand how you got 1/2. Not that it is important, but if you'd like to discuss it, I'd be more than happy to clarify that as well, if it isn't clear.

Answer 11

OH! I see, I got \[(\lim \theta \rightarrow0 \sin 5\theta/\theta)\div(1+\lim \theta \rightarrow0 \sin8\theta/\theta) \times (\lim \theta \rightarrow0 1/\cos8\theta)\]
And then from there I got that 1/(1+ (1x1))....since I thought that they all approached one as theta approached zero.

Answer 12

What a silly mistake :p