Isn't this problem about continuous functions in my book wrong?

Question

The problem:

If $f(x)=\begin{cases}\frac{\sin^2(ax)}{x^2}&x\neq0\\ 1&x=0\end{cases}$, show that $f(x)$ is discontinuous at $x=0$. If $f(x)$ is continuous at $x=0$, prove that $a=1$.

My attempt:

$$\lim_{x\to 0^{-}}\frac{\sin^2ax}{x^2}=\lim_{x\to 0^{+}}\frac{\sin^2ax}{x^2}\neq f(0)$$

So, the function is discontinuous at $x=0$. If it is continuous at $x=0$, then

$$\lim_{x\to 0}\frac{\sin^2ax}{x^2}=f(0)$$

$$\lim_{x\to 0}\frac{\sin^2ax}{x^2}=1\tag{1}$$

My question:

1. I could've easily proved that $a=1$ from $(1)$ if the question had $\frac{\sin ax}{x^2}$ instead of $\frac{\sin^2ax}{x^2}$. So, isn't the problem wrong?

Answer 1

We know that:

$$\lim_{x\to0}\frac{\sin^2ax}{x^2}=\lim_{ax\to0}a^2\cdot\frac{\sin^2ax}{(ax)^2}=a^2$$

Which gives that $a=\pm1$ for $f$ to be continuous at $0$, as $f(0)$ was defined to be $1$.

However you wrote:

I could’ve easily proved $a=1$ if we had $\frac{\sin ax}{x^2}$

But this function is continuous at zero if and only if $a=0$, because the $x^2$ is not matched by a $\sin^2$ and we get a pole at zero for non-zero $a$:

More formally, the limit is:

$$\left(\lim_{x\to0}\frac{\sin ax}{x}\right)\cdot\left(\lim_{x\to0}\frac{1}{x}\right)$$

Which is asymptotic when $a\neq0$.

Answer 2

Actually, $\displaystyle\lim_{x\to0}\frac{\sin^2(ax)}{x^2}=a^2$. Therefore, $f$ is continuous at $0$ if and only if $a=\pm1$.