Does the Limit of g(x) Equal the Square Root of the Limit of f(x)?

[Felafel]

Homework Statement

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 $\displaystyle \lim_{x \to x_0} g(x)=\sqrt{\displaystyle \lim_{x \to x_0} f(x)}$

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

 

[pasmith]

Homework Statement

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 $\displaystyle \lim_{x \to x_0} g(x)=\sqrt{\displaystyle \lim_{x \to x_0} f(x)}$

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

Sadly this is not sufficient.

Your aim is to prove:
(\forall \epsilon &gt; 0)(\exists \delta &gt; 0)(\forall x \in A)(|x - x_0| &lt; \delta \Rightarrow |g(x) - \sqrt L|&lt; \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 &gt; 0 there exists \eta &gt; 0 such that for all f(x) \geq 0, if |f(x) - L| &lt; \eta then |g(x) - \sqrt L | &lt; \epsilon.

The second way is to let \epsilon &gt; 0 be arbitrary, start from
|g(x) - \sqrt L| = \left|\sqrt{f(x)} - \sqrt L\right|&lt; \epsilon
and multiply by \sqrt{f(x)} + \sqrt L to get
<br /> |f(x) - L| - \epsilon\left(\sqrt{f(x)} + \sqrt L\right) &lt; 0<br />
In each of the cases f(x) &gt; L and f(x) &lt; 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.)

 

[SammyS]

I fixed some obvious typos (in red).

Homework Statement

given
$A \subset \mathbb{R}$

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

considering the function g such that:

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

The Attempt at a Solution

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

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 .

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