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$\lim_{x \to c} ((f(x))^n)$ = $(\lim_{x \to c} (f(x)))^n$

I have proved this for natural numbers, $n=0$, integers, and even rational numbers. How does one prove this property for irrational numbers, and therefore for real numbers, assuming the proofs for the above have been completed.

I have been proving the properties using the delta epsilon definition of a limit.

Laksh
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1 Answers1

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As stated, the result is not true. To be precise, if $\lim_{x\rightarrow c}f(x)$ exists, then your statement is true.

For example, if you have a function $f$ that is $1$ at every rational number and $-1$ at every irrational number, then $\lim_{x\rightarrow c}(f(x))^{2}=1$ exists, but $(\lim_{x\rightarrow c}f(x))^{2}$ does not exist because the inner limit does not exist.

More generally, if $f$ is a function for which $\lim_{x\rightarrow c}f(x)=L$ exists, and if $g$ is a function defined on an open neighborhood of $L$ that is continuous at $L$, then $\lim_{x\rightarrow c}g(f(x))$ exists and equals $f(\lim_{x\rightarrow c}f(x))$. Sometimes the general is easier to prove than the specific.

Disintegrating By Parts
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  • So basically let the g function be the "to the power of n" function? If that was so wouldn't you also have to prove it is continuous at L? Sorry I'm a bit confused. – Laksh Jun 25 '15 at 05:14
  • Even so, specifically how would you prove my question given that the limit exists? – Laksh Jun 25 '15 at 05:15
  • @Aced125 : It's automatic if you know that $x \mapsto x^{n}$ is a continuous function. Use $x^{n} =e^{n\ln x}$. – Disintegrating By Parts Jun 25 '15 at 07:46