Question

Question #dd6b9

Answer 1

Because `lim_(xrarr0)cos5x = cos 0 =1`, the challenge is finding the limit of the numerator.

Explanation:

We should know that

`lim_(thetararr0)sintheta/theta = 1` and `lim_(thetararr0)(1-costheta)/theta=0`

(We don't need the second, but these two limits "go together".

Now we need to remember some trigonometry.

How else can we write `csc2x`?

Eventually we recall `csc2x = 1/(sin2x)`.

Ok, so `x csc2x = x/(sin2x)`.

We want theta to be `2x`, so multiply by `1` in the form `2/2`

`x csc2x = 1/2 overbrace((2x)^theta)/(sin underbrace(2x)_theta)`.

The limit as `xrarr0` is `1/2*1 = 1/2` .