$\lim_{x\to0}\frac{\sin^2x}{3x^2}$.

Question

I would like to compute $\lim_{x\to0}\frac{\sin^2x}{3x^2}$. In the book it is given that $\lim_{x\to a}g(x)f(x) = \lim_{x\to a}g(x)\lim_{x\to a}f(x)$. I was going to use this idea and write $\lim_{x\to0}\frac{\sin^2x}{3x^2}=\lim_{x\to0}\frac{\sin x}{3x}\frac{\sin x}{3x}=1/3*1/3=1/9$. However, the answer is $1/3$. Now my question is the following: Why can't I use the product formula and how to get the desired answer?

Answer 1

You are almost right:$$\lim_{x\to0}\frac{\sin^2x}{3x^2}=\frac13\left(\lim_{x\to0}\frac{\sin x}x\right)\left(\lim_{x\to0}\frac{\sin x}x\right)=\frac13\times1\times1=\frac13.$$

Answer 2

You were really close!

Your one error is that $\displaystyle \frac{\sin(x)}{3x} \cdot \frac{\sin(x)}{3x}=\frac{\sin(x)^2}{9x^2}$, not $\displaystyle \frac{\sin(x)^2}{3x^2}$.

So, instead, you could just multiply the limits of $\displaystyle \frac{\sin(x)}{3x}$ and $\displaystyle \frac{\sin(x)}{x}$.