Question

lim(x->0) (sin^2(2 x))\/x^2

Answer 1

\[\lim_{x \rightarrow 0}\frac{ \sin ^{2}(2x) }{x^2 }\]

Answer 2

Seems too much to hope, but is L'Hôpital allowed?

Answer 3

Sadly no :(

Answer 4

Can use Sinx/x = 1 and 1-cosx/x = 0

Answer 5

Well in that case, why don't we try and decompose the function into something that resembles those two forms?

Answer 6

How about this:
\[\Large \frac{\sin^2(2x)}{x^2}= \frac{\sin(2x)}{x}\cdot\frac{\sin(2x)}{x}\]

Answer 7

I was thinking
\[\frac{ (Sin2x)(Sin2x) }{x^2 }*\frac{ 2 }{ 2 }\]
Does this make sense?

Answer 8

Then I come up with 2 as Sin2x/2x =1

Answer 9

But http://www.wolframalpha.com/widgets/view.jsp?id=265eceb6d4d961057f1b483a558e2885

Says its 4

Answer 10

But you need two of them.

Answer 11

Ohhhh

Answer 12

\[\Large \frac{\sin^2(2x)}{x^2}= \frac{\sin(2x)}{\color{red}x}\cdot\frac{\sin(2x)}{\color{green}x}\]

Answer 13

\[\frac{ \sin^2(2x) }{ x^2 }*\frac{ 2 }{ 2 }*\frac{ 2 }{ 2 } = 4\]

Answer 14

Bingo :P

Answer 15

Am I right?

Answer 16

Nice <3

Answer 17

Though, allow me this one little nitpick:
\[\Large \color{red}{\lim_{x\rightarrow0}}\frac{ \sin^2(2x) }{ x^2 }*\frac{ 2 }{ 2 }*\frac{ 2 }{ 2 } = 4\]

Answer 18

Yup, perfect.