$\lim_{x\to a}(f(x)+\frac{1}{|f(x)|})=0$. Find $\lim_{x\to a}f(x)$

Question

Let a function $f$ be defined in a hollow neighborhood of $a\in \mathbb{R}$, and suppose : $$\lim_{x\to a}\left(f(x)+\frac{1}{|f(x)|}\right)=0$$

Find $\lim\limits_{x\to a}f(x)$ and prove that this is the limit by definition.

My approach: It is easy to see that if this limit exists it should be $-1$ but I can't prove it by definition. Please help.

Answer 1

Note that if $\lim_{x\to a}(f(x)+\frac{1}{|f(x)|})=0$, then for $x$ close enough to $a$, $f(x)+\frac{1}{|f(x)|}<1$. A quick look at the graph of $x+\frac{1}{|x|}$ shows that this means $f(x)$ is negative, so you actually have that $\lim_{x\to a}(f(x)-\frac{1}{f(x)})=0$.

Some algebra shows that $\lim_{x\to a}(f(x)-\frac{1}{f(x)})^2+4=\lim_{x\to a}(f(x)+\frac{1}{f(x)})^2=4$, and thus that $\lim_{x\to a}(f(x)+\frac{1}{f(x)})=\pm2$. Noting that $f$ is negative as $x\to a$, we can take the negative root, thus seeing that:

$$\lim_{x\to a}(f(x)-\frac{1}{f(x)})=0$$

$$\lim_{x\to a}(f(x)+\frac{1}{f(x)})=-2$$

and thus, finally,that $f(x)\to -1$ as $x \to a$.

Note: One can use the fact that $g(x)=x+\frac{1}{|x|}$ has a continuous inverse around $0$ directly, though this arguably presupposes extra knowledge about inverse functions. In any case, if you're comfortable using it, it provides a slightly quicker route to the answer.