Question

Help would be greatly appreciated:
lim
x→0 (cos 9x − 1)/(sin 8x)

Answer 1

Can you use l'hospital's rule?

Answer 2

What is that?

Answer 3

If you don't know what it is then I assume you are suppose to do this algebraically.

Answer 4

Or using trig identities.

Answer 5

Or both.

Answer 6

But it does say if you plug in what x is approaching and you have 0/0 and infinity/ infinity then finding the limit to lim (x->whatevers) (f'/g') =lim (x -> whatevers) (f/g)

Answer 7

Haha im kind of confused..

Answer 8

We need to use some kinda of trig identity/ algebraic manipulation so when we plug in 0 we don't get 0 on bottom.

Answer 9

well we know that
\[\lim_{x \rightarrow 0} \frac{\sin(x)}{x}=1 ; \lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}\]

Answer 10

Take your problem and divide top and bottom by 8x like this:
\[\lim_{x \rightarrow 0}\frac{\cos(9x)-1}{1} \cdot \frac{1}{\sin(8x)}\]
\[\lim_{x \rightarrow 0}\frac{\cos(9x)-1}{1} \cdot \frac{8x}{\sin(8x)} \cdot \frac{1}{8x}\]
Now since we can also use that one limit. Divide and multiply by 9x.

Answer 11

\[\lim_{x \rightarrow 0}\frac{\cos(9x)-1}{9x} \cdot \frac{8x}{\sin(8x)} \cdot \frac{9x}{8x}\]
That should help.
Evaluate the following three limits:
\[\lim_{x \rightarrow 0}\frac{\cos(9x)-1}{9x} \]
\[\lim_{x \rightarrow 0}\frac{8x}{\sin(8x)}\]
\[\lim_{x \rightarrow 0}\frac{9x}{8x}\]

Answer 12

How do you know cosx-1/x = sinx/x?

Answer 13

Those don't equal
as x->0, sin(x)/x goes to 1 by the squeeze thm
we can use that along with some algebra to show as x->0, (cos(x)-1)/x goes to 0.

Answer 14

Could you explain that further? Im not sure I udnerstand

Answer 15

The squeeze thm?

Answer 16

also called the sandwich thm

Answer 17

I think there is another name for it but can't think of it.

Answer 18

Yes but how do you use that to prove cosx-1?

Answer 19

\[\lim_{x \rightarrow 0} \frac{\cos(x)-1}{x} \cdot \frac{\cos(x)+1}{\cos(x)+1}=\lim_{x \rightarrow 0}\frac{-\sin^2(x)}{x(\cos(x)+1)}\]
\[=\lim_{x \rightarrow 0}\frac{\sin(x)}{x} \cdot \frac{-\sin(x)}{\cos(x)+1}=1 \cdot \frac{-\sin(0)}{\cos(0)+1}=1 \cdot \frac{0}{2}=1(0)=0\]

Answer 20

The squeeze thm states something like this
if g<f<h where g,h,f are all continuous around x=a (doesn't necessarily have to be continuous at x=a)
and\[\lim_{x \rightarrow a}g=\lim_{x \rightarrow a}h =L \text{ then } \lim_{x \rightarrow a}f=L\]