[SPLIT] find limit, x -> 0, of sin^2(2x) / sin^2(3x)

[uniqueownz]

i have more one though its quite confusing.

lim sin[sup:1juovh13]2[/sup:1juovh13] 2x
[sup:1juovh13]x->0[/sup:1juovh13] sin[sup:1juovh13]2[/sup:1juovh13] 3x

 

[skeeter]

this may help ...

\(\displaystyle \frac{4}{9} \cdot \lim_{x \to 0} \frac{9x^2}{4x^2} \cdot \frac{\sin^2(2x)}{\sin^2(3x)}\)

 

[uniqueownz]

mhmmm dont kno how to use that

 

[skeeter]

what do you know about this limit?

\(\displaystyle \lim_{x \to 0} \frac{\sin{x}}{x}\)

 

[galactus]

Did you try expanding it out?.

\(\displaystyle \frac{sin^{2}(2x)}{sin^{2}(3x)}=\frac{4cos^{2}(x)}{16cos^{4}(x)-8cos^{2}(x)+1}\)

Now, can you see the limit?.

 

[uniqueownz]

yea that equals 1 skeeter

 

[skeeter]

good. so, what about ...

\(\displaystyle \lim_{x \to 0} \frac{\sin(ax)}{ax}\)

???

 

[uniqueownz]

that is also equal to 1 skeeter

 

[skeeter]

good! what about ...

\(\displaystyle \lim_{x \to 0} \frac{bx}{\sin(bx)}\)

???

 

[uniqueownz]

again, that is also equal to 1

 

[skeeter]

take another look at this ...

\(\displaystyle \frac{4}{9} \cdot \lim_{x \to 0} \frac{9x^2}{4x^2} \cdot \frac{\sin^2(2x)}{\sin^2(3x)}\)

remember that \(\displaystyle 9x^2 = (3x)^2\) and \(\displaystyle 4x^2 = (2x)^2\)

 

[uniqueownz]

so does the top and bottom just like cancel when u multiply by 9x[sup:3e5ym8s7]2[/sup:3e5ym8s7] / 4x[sup:3e5ym8s7]2[/sup:3e5ym8s7] and the answer is 4/9?

 

[skeeter]

so does the top and bottom just like cancel when u multiply by 9x[sup:1r2xkr0o]2[/sup:1r2xkr0o] / 4x[sup:1r2xkr0o]2[/sup:1r2xkr0o] and the answer is 4/9?

yes, the limit is 4/9, but no, they do not "cancel".
\(\displaystyle \frac{4}{9} \cdot \frac{9x^2}{4x^2} = 1\)
when you multiply this by \(\displaystyle \frac{\sin^2(2x)}{\sin^2(3x)}\), it does not change its value.

rearrange what you have ...

\(\displaystyle \frac{4}{9} \lim_{x \to 0} \frac{\sin^2(2x)}{(2x)^2} \cdot \frac{(3x)^2}{\sin^2(3x)} =\)

\(\displaystyle \frac{4}{9} \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \frac{\sin(2x)}{2x} \cdot \frac{3x}{\sin(3x)} \cdot \frac{3x}{\sin(3x)}=\)

\(\displaystyle \frac{4}{9} \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \lim_{x \to 0} \frac{3x}{\sin(3x)} \cdot \lim_{x \to 0} \frac{3x}{\sin(3x)}= \frac{4}{9} \cdot 1 \cdot 1\cdot 1\cdot 1 = \frac{4}{9}\)

 

[uniqueownz]

thank you soo much. i understand it now. thank you!

 

[galactus]

\(\displaystyle \lim_{x\to 0}\left(\frac{sin(2x)}{sin(3x)}\right)^{2}\)

Inside the parentheses, divide top and bottom by x.

\(\displaystyle \lim_{x\to 0}\left(\frac{\frac{sin(2x)}{x}}{\frac{sin(3x)}{x}}\right)^{2}\)

\(\displaystyle \lim_{x\to 0}\left(\frac{\frac{2sin(2x)}{2x}}{\frac{3sin(3x)}{3x}}\right)^{2}\)

Now, we can see from the famous limit previously discussed that \(\displaystyle \lim_{x\to 0}\frac{sin(2x)}{2x}=1 \;\ and \;\ \lim_{x\to 0}\frac{sin(3x)}{3x}=1\)

So, we get \(\displaystyle \left(\frac{2}{3}\right)^{2}=\frac{4}{9}\)