# Prove existence of a limit

1. Jan 13, 2013

### Felafel

1. The problem statement, all variables and given/known data

given
$A \subset \mathbb{R}$

$f:A \subset \mathbb{R} \to \mathbb{R}^+$

considering the function g such that:

$g(x):=\sqrt{f(x)} x \in A$ with $x_0$ limit point in A.
Prove that if $\displaystyle \lim_{x \to x_0} f(x)$ exists, then $\displaystyle \lim_{x \to x_0} g(x)=\sqrt{\displaystyle \lim_{x \to x_0} f(x)}$

3. The attempt at a solution

Being $\displaystyle \lim_{x \to x_0}=L$
By definition of limit:
$\forall \epsilon>0 \exists \delta >0: \forall x: 0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon$

Let: $\epsilon=\epsilon |\sqrt{f(x)}+\sqrt L|$

$|f(x)-L|<\epsilon |\sqrt{f(x)}+\sqrt L| \Rightarrow |\frac{f(x)-L}{\sqrt{f(x)}+\sqrt L}|< \epsilon$ $\Rightarrow |\sqrt{f(x)}+\sqrt L}|<\epsilon$

is it sufficient? thank you in advance

2. Jan 13, 2013

### pasmith

$$(\forall \epsilon > 0)(\exists \delta > 0)(\forall x \in A)(|x - x_0| < \delta \Rightarrow |g(x) - \sqrt L|< \epsilon).$$

There are two ways to do this. The easiest is to use the fact that the square root function is continuous, which tells you that for all $\epsilon > 0$ there exists $\eta > 0$ such that for all $f(x) \geq 0$, if $|f(x) - L| < \eta$ then $|g(x) - \sqrt L | < \epsilon$.

The second way is to let $\epsilon > 0$ be arbitrary, start from
$$|g(x) - \sqrt L| = \left|\sqrt{f(x)} - \sqrt L\right|< \epsilon$$
and multiply by $\sqrt{f(x)} + \sqrt L$ to get
$$|f(x) - L| - \epsilon\left(\sqrt{f(x)} + \sqrt L\right) < 0$$
In each of the cases $f(x) > L$ and $f(x) < L$ the left hand side is a quadratic in $\sqrt{f(x)}$, so you can derive conditions on $\sqrt{f(x)}$ for the inequality to hold. You should be able to rearrange these to get an upper bound on $|f(x) - L|$ in terms of $\epsilon$ and $L$.

(What you're actually doing here is proving that square root is continuous and using a slightly cumbersome notation in order to do so.)

3. Jan 13, 2013

### SammyS

Staff Emeritus
I fixed some obvious typos (in red).

From the above, you know that given any ε, you can find a δ that works for f(x).

Now, what you need to show is that for any (arbitrary) ε>0, you can find a δ which works for g(x).

You don't get to choose an ε, which is what you have below. Also, when finding a δ, it won't depend on x. I can possibly depend on x0, but not on x .