The Limit of a Root of a Function is the Root of that Limit (delta-epsilon)

[5hassay]

Homework Statement

Prove:

If $\lim_{x \rightarrow a} f \left( x \right) = l$, then, if $\sqrt[n]{f \left( x \right)}$ exists, $\lim_{x \rightarrow a} \sqrt[n]{f \left( x \right)} = \sqrt[n]{l}$.

Homework Equations

Nothing really...

The Attempt at a Solution

Okay, so here goes my attempt...

Proof. We can assume that

$\forall \varepsilon > 0$, $\exists \delta_1 > 0$ : $\forall x$, $0 < \left| x - a \right| < \delta_1 \Longrightarrow \left| f \left( x \right) - l \right| < \varepsilon$,​

whereas, we want to show

$\forall \varepsilon > 0$, $\exists \delta_2 > 0$ : $\forall x$, $0 < \left| x - a \right| < \delta_2 \Longrightarrow \left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right| < \varepsilon$.​

But, $\varepsilon > \left| f \left( x \right) - l \right| \geq \left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right|$. Therefore, we conclude that given $\varepsilon > 0$, if, for all $x$, $0 < \left| x - a \right| < \delta$ for some such $\delta$, then we have both $\left| f \left( x \right) - l \right| < \varepsilon$ and $\left| \sqrt[n]{f \left( x \right)} - \sqrt[n]{l} \right| < \varepsilon$.

QED

So, I guess the first remark would be if my method was even right, where the second would be if my inequality is true (I can't think of a counterexample).

Anyway, much appreciation for any help.

 

[vela]

For n=2, f(x) = 1/2 and l = 0 is a counterexample.

 

[5hassay]

For n=2, f(x) = 1/2 and l = 0 is a counterexample.

Oh, drat. Then, I am at a loss on what to do.

Hmm... I'm likely misunderstanding, but if $f \left( x \right) = 1/2$, then wouldn't $l = \lim_{x \rightarrow a} 1/2 = 1/2$ for any $a$? That is, I don't understand how the limit $l$ could be $0$.

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Much appreciation

 

[vela]

Sorry, what I wrote wasn't very clear. I meant if f=1/2 for some value of x near a, then the implication your proof relied on doesn't hold. I didn't mean that f was a constant function.

 

[JG89]

$|\sqrt{f(x)} - \sqrt{L}| = \frac{f(x) - L}{|\sqrt{f(x)} + \sqrt{L}|}$.

The denominator of the right side of the equation can never be zero unless f = 0 in some neighborhood of x= a and L = 0 (in which case the proposition you're trying to prove is trivial). Why?

Once you answer that, let x approach a and see what the limit of the right side of the equation is.

 

[5hassay]

Sorry, what I wrote wasn't very clear. I meant if f=1/2 for some value of x near a, then the implication your proof relied on doesn't hold. I didn't mean that f was a constant function.

$|\sqrt{f(x)} - \sqrt{L}| = \frac{f(x) - L}{|\sqrt{f(x)} + \sqrt{L}|}$.

The denominator of the right side of the equation can never be zero unless f = 0 in some neighborhood of x= a and L = 0 (in which case the proposition you're trying to prove is trivial). Why?

Once you answer that, let x approach a and see what the limit of the right side of the equation is.

Alright... I've been trying to understand these hints, but I unfortunately don't seem to get anywhere. However, I realized that I don't need this proof as much as I thought, so I am just going to put this on the 'back burner' (if I got that right) for now. Thanks for the help!