(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Prove:

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

2. Relevant equations

Nothing really...

3. The attempt at a solution

Okay, so here goes my attempt...

Proof.We can assume that

[itex] \forall \varepsilon > 0 [/itex], [itex] \exists \delta_1 > 0 [/itex] : [itex] \forall x [/itex], [itex] 0 < \left| x - a \right| < \delta_1 \Longrightarrow \left| f \left( x \right) - l \right| < \varepsilon [/itex],

whereas, we want to show

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

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

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.

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

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