prove limit of function by definition Let $a\in\mathbb R $. Assume that:
$$
\lim _{x\to a} \left ( f(x) +\frac{1}{|f(x)|}  \right ) = 0
$$
Find $\lim _{x\to a} f(x) $ and prove by definition that it is the limit.
I am guessing that the limit is $-1$ but I have difficulty to show that. Any suggestions? Thanks for helpers!
 A: Look at the function $\phi(t) = t + {1 \over |t|}$.
Note that if $t >0$ then $\phi(t) \ge 2$.
Note that $\lim_{t \to 0} \phi(t) = \infty$.
In particular, note that $\phi(t) \le {3 \over 2}$ iff $t \le -{1 \over 2}$.
Note that when restricted to $(-\infty,-{1 \over 2}]$, $\phi$ is invertible and the inverse $\eta = \phi^{-1}$ is continuous.
Note that if $y \le {3 \over 2}$ and $\phi(t) = y$, then we must have $t \le -{1 \over 2}$. Hence $\phi(t) = t -{1 \over t} = y$ and
multiplying across by $t$ gives the equation $t^2-t y -1 = 0$ and
hence $t = { 1\over 2} (y-\sqrt{y^2+4})$. Hence $\eta(y) = { 1\over 2} (y-\sqrt{y^2+4})$, which is continuous.
You are given that $\lim_{x \to a} \phi(f(x)) = 0$, in particular,
for some $\delta>0$ we have $\phi(f(x)) < {3 \over 2}$
for all $x \in B(a,\delta)$ and so $f(x) < -{1 \over 2}$ for all
$x \in B(a,\delta)$. (In particular, for $x \in B(a,\delta)$, we have $\eta(\phi(f(x))) = f(x)$).
Since $\eta$ is continuous at $0$ we have
$$-1 = \eta(0) = \eta(\lim_{x \to a} \phi(f(x))) = \lim_{x \to a} \eta(\phi(f(x))) = \lim_{x \to a} f(x)$$
