Question

how to solve: lim (x-->0) sin(2x)sin(3x)/sin(3x^2)

Answer 1

it is look like this
\[\large \lim_{x \rightarrow 0}\frac{\sin(2x)\sin(3x)}{\sin(3x ^{2})}\]Is this your question?

Answer 2

yes

Answer 3

We know that
\[\lim_{x\to 0} \frac{ \sin x}{x}=1-------------------(1)\]
and also
\[\lim_{x\to 0} \frac{ \sin x^2}{x^2}=1-------------------(2)\]
so her we have
\[\lim_{x\to 0} \frac{ \sin 2x\sin 3x }{\sin 3x^2 }\]
We have to create forms like (1) and (2) to evaluate the limit
Dividing and mulitplying by \( 2x \times 3x\)
\[\lim_{x\to 0} \frac{ \sin 2x\sin 3x \times 2x \times 3x }{\sin 3x^2 \times 2x \times 3x
}=1\]
So we have now
\[\lim_{x\to 0} \frac{\sin 2x }{2x} \times \frac{ \sin 3x}{3x} \times \frac{ 2x \times 3x}{ \sin 3x^2}\]
\[\lim_{x\to 0} \cancel {\frac{\sin 2x }{2x}} 1 \times \cancel{\frac{ \sin 3x}{3x}} 1 \times 2\frac{ 3x^2}{ \sin 3x^2}\]
Now we have
\[\lim _{x\to 0} 2\cancel{\frac{ 3x^2}{ \sin 3x^2}}1\]
so we get
\[\lim_{x\to 0} \frac{ \sin 2x\sin 3x }{\sin 3x^2 }=2\]

Answer 4

thank you!

Answer 5

@azalea Welcome to Open Study:)