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I need help to verify the following

Prove that if does not equal 0

lim of x approaches a: x^-n = a^-n

I know to prove lim of x approaches a: x^n = a^n requires induction so I believe that this problem requires the same.

lim of x approaches a: [f(x)]^-n ... = k^-n

inductive step

lim of x approaches a: [f(x)]^-n-1 * lim of x approaches a: [f(x)] =

k^-n-1*k = k

Does this prove the question?

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1 Answer 1

up vote 1 down vote accepted

If the statement is known to you with positive exponents (as you say), then you can prove the statement for negative ones by using that the function $x\mapsto\dfrac1x$ is continuous.

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So since I have already proved lim x to a: x^n by induction previously and all that, I can say lim x to a: x^-n is true because x > 1/x is continuous? –  Benji_Bombadill Oct 16 '13 at 13:23
    
Yes, that's what I'm saying. –  Rasmus Oct 16 '13 at 13:32

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