Proving one sided limits

Question

I would like to know if my solution is correct for the following limit.

$\displaystyle\lim_{x\to0^+}\frac{x+2}{\sqrt{x}}$

We claim the limit is $\infty$.

For any $\alpha\in\mathbb{R}$, there exists $\delta > 0$ such that for any $x\in (0,\delta)$ we have $x > 0$, $x < \delta$.

Thus $\sqrt{x} < \sqrt{\delta}$ and so $\dfrac{1}{\sqrt{x}} > \dfrac{1}{\sqrt{\delta}}$. Furthermore, since $x > 0$ we have $x + 2 > 2$.

Thus, from $f(x) = \dfrac{x+2}{\sqrt{x}}$ it follows $f(x) > \dfrac{2}{\sqrt{\delta}}$

Let $\delta = \dfrac{4}{\alpha^2}$

$f(x) > \dfrac{2}{\sqrt{4/\alpha^2}}$

$f(x) > \dfrac{2\alpha}{2}$

$f(x) > \alpha$

Therefore, $\lim_{x\to 0^+} f(x) = \infty$ by definition.

If it is not correct, what do I need to change?

Answer 1

Correct, except that the part

For any $\alpha\in\mathbb{R}$, there exists $\delta > 0$ such that for any $x\in (0,\delta)$ we have $x > 0$, $x < \delta$.

is true, but meaningless. Substitute it with

Let $\delta>0$ and suppose $x\in(0,\delta)$.

On the other hand, it's much easier. Let $\alpha>0$. Then, if $0<x<1/\alpha^2$, we have $$ \frac{x+2}{\sqrt{x}}>\frac{1}{\sqrt{x}}>\alpha $$