Finding Limits of Trig Functions

[cheerspens]

Homework Statement

The first problem I'm having difficulty with is

$\stackrel{lim}{x\rightarrow0}$ $\frac{sin x}{5x}$

And the second is:
$\stackrel{lim}{x\rightarrow0}$ $\frac{sin x(1-cos x)}{2x^{2}}$

Homework Equations

I assume that for the first problem I need to simplify it to the rule where $\stackrel{lim}{x\rightarrow0}$ $\frac{sin x}{x}$=1
and the second would probably need to simplify to follow the rule $\stackrel{lim}{x\rightarrow0}$ $\frac{1-cos x}{x}$=0

The Attempt at a Solution

What I mainly need help with is how to get started. For the first problem, how do I get rid of the 5x at the bottom?
For the second problem should I square the entire thing and end up with $\stackrel{lim}{x\rightarrow0}$ $\frac{2sin x}{4x^{4}}$ then go from there or is that even correct?

 

[rock.freak667]

The Attempt at a Solution

What I mainly need help with is how to get started. For the first problem, how do I get rid of the 5x at the bottom?
For the second problem should I square the entire thing and end up with $\stackrel{lim}{x\rightarrow0}$ $\frac{2sin x}{4x^{4}}$ then go from there or is that even correct?

Rewrite the first one as (1/5)(sinx/x) then take the limit.

use the fact that x²=x*x and then try to get sinx/x and (1-cosx)/x then use the fact that lim(x→a) f*g = lim(x→a) f * lim(x→a) g

 

[rude man]

Use the power series for sin(x) and cos(x).

You already know the first answer is 1/5.

Second problem:
cos x = 1 - ax^2 + bx^4 + ... etc. since it's an even function.
sin x = cx + dx^3 + ... etc. since it's an odd function.
Then take the first two terms of cos x and the first term of sin x in your expression, realize that the lowest power of x in the numerator is higher than the highest power of x in the denominator, and bingo!