Given $\lim_{x \rightarrow a}f(x)+\frac1{\left|f(x)\right|}=0$, what is $\lim_{x \rightarrow a}f(x)$? Another question I am struggling with:
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be defined in a neighborhood of $a$ (except maybe $a$ itself) and suppose that, $\lim_{x \rightarrow a}\left(f(x)+\dfrac{1}{\left|f(x)\right|}\right)=0$. Find $\lim_{x \rightarrow a}f(x)$, and prove by definition, that this is indeed the limit.

Thank you all for your answers!!
But.. How can I prove by definition that this is indeed the limit? take $\epsilon>0$, I have to show that there exists a $\delta>0$ such that, $|x-a| \leq \delta$ implies $|f(x)+1| \leq \epsilon$. now, I do know that there exists an $\delta_1>0$, such that $|f(x)+\frac{1}{|f(x)|}|< \epsilon$.Or, $\frac{\epsilon}{k}$ for some constant $k$ But, how can I continue from there?
 A: If $f(x)$ tends to a limit as $x \to a$ then this limit must be $-1$. First we note that the limit $\lim_{x \to a}f(x)$, if it exists, must be negative. If the limit were positive then both $f(x)$ and $1/|f(x)|$ would have to be positive as $x \to a$ and since their sum tends to $0$ each must tend to $0$ which is impossible. Similarly the limit can't be $0$ otherwise $1/|f(x)|$ would tend to $\infty$. Now if we have the limit as $L < 0$, then $|f(x)| \to -L$ and we get $L - (1/L) = 0$. This is possible only if $L = -1$.
A: Graphing the auxiliary function $$\phi:\quad u\mapsto u+{1\over |u|}\qquad(u\in\dot{\mathbb R})$$
we see that (a) $\ |\phi(u)|\geq{5\over6}$ when $u\leq-{3\over2}$ or $u\geq-{1\over2}$, and that (b) the function $\phi$ is continuous and strictly increasing on the interval $I:\ -{3\over2}<u<-{1\over2}$. From (b) we conclude that $\phi\restriction I$ has an inverse $\phi^{-1}$  which is continuous on $\phi(I)$.
Now we are told that the function
$$g(x):=f(x)+{1\over |f(x)|}=\phi\bigl(f(x)\bigr)$$
has limit $0$ when $x\to a$. It follows that there is a neighborhood $U$ of $a$ such that $|g(x)|<{5\over6}$ for all $x\in\dot U$. According to (a) above this enforces $-{3\over2}<f(x)<-{1\over2}$ for all $x\in\dot U$. It follows that $f(x)=\phi^{-1}\bigl(g(x)\bigr)$ for all $x\in\dot U$, so that we obtain
$$\lim_{x\to a}f(x)=\lim_{x\to a}\left(\phi^{-1}\bigl(g(x)\bigr)\right)=\phi^{-1}(0)=-1\ .$$
