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Answer 1 · 2017-01-17T18:23:36

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

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 $1/2*1 = 1/2$ .