Limit of a trig function

[peblez]

lim x ---> 0

sin ^2 3x / 4 x^2

i did

3/4 sin ^2 x / x ^ 2

= 3/4


but answer says 9/4

 

[tkhunny]

It was rather magic how you pulled that '3' out of the argument. Put it back and rethink it.

What's this limit? sin(x)/x as x --> 0?

How about this one? sin(3x)/x as x --> 0

This one? ([sin(3x)]^2)/[x^2] as x --> 0

 

[peblez]

i seee.

so basically when i pulled out the three, i needed to square it first?

 

[soroban]

Hey, peblez!

Stop it! . . . \(\displaystyle \sin(3x) \text{ is }not\text{ equal to }3\sin(x)\)

\(\displaystyle \L\lim_{x\to0}\frac{\sin^2(3x)}{4x^2}\)

You're expected to use: \(\displaystyle \L\:\lim_{\theta\to0}\frac{\sin\theta}{\theta}\:=\:1\)

. . and we have to hammer the problem into that form.


\(\displaystyle \text{Multiply by }\frac{9}{9}:\L\;\;\frac{9}{9}\cdot\frac{\sin^2(3x)}{4x^2} \;=\;\frac{9}{4}\cdot\frac{\sin^2(3x)}{9x^2} \;=\;\frac{9}{4}\cdot\left[\frac{\sin(3x)}{3x}\right]^2\)


If \(\displaystyle x\to0\), we see that: \(\displaystyle 3x\to0\)

So we have: \(\displaystyle \L\:\lim_{3x\to0}\frac{9}{4}\cdot\left[\frac{\sin(3x)}{3x}\right]^2 \;=\;\frac{9}{4}\cdot\lim_{3x\to0}\left[\frac{\sin(3x)}{3x}\right]^2\)

. . \(\displaystyle \L=\;\frac{9}{4}\cdot\left[\lim_{3x\to0}\frac{\sin(3x)}{3x}\right]^2 \;=\;\frac{9}{4}\cdot[1^2] \;=\;\frac{9}{4}\)