Show that if $\lim_{x\to c}(f(x))^2=0$ then $\lim_{x\to c}f(x)=0$

Question

Let $c\in\mathbb{R}$ and let $f:\mathbb{R}\to\mathbb{R}$ be such that $$\lim_{x\to c}(f(x))^2=L$$ Show that if $L=0$, then $$\lim_{x\to c}f(x)=0$$

Using the $\delta$ - $\varepsilon$ definition, it is given that:

$\forall \varepsilon>0$ $\exists\delta_1>0$ such that, if $|x-c|<\delta_1$ then, $|f^2(x)|<\varepsilon$

Using this, what we need to show is that:

$\forall \varepsilon>0$ $\exists\delta_2>0$ such that, if $|x-c|<\delta_2$ then, $|f(x)|<\varepsilon$

My approach:

Since, $\varepsilon$ is taken to be small, we can say that $\varepsilon<1$. I take $\delta_2=\delta_1$ $\forall\varepsilon>0$.
$\Rightarrow|f^2(x)|<|f(x)|<\varepsilon<1$, which doesn't get me anywhere.

(In fact, this makes it look like that the converse is true!)

My approach is obviously very wrong. How should I actually proceed?

Answer 1

Let $\varepsilon>0$. You know that there is a $\delta>0$ such that $$\lvert x\rvert<\delta\implies\bigl\lvert f^2(x)\bigr\rvert<\varepsilon^2.$$But$$\bigl\lvert f^2(x)\bigr\rvert<\varepsilon^2\iff\bigl\lvert f(x)\bigr\rvert<\varepsilon.$$