Prove that $$\lim_{x\mathop\to a}x^n=a^n$$
Is there an easy and rigorous way to show this using epsilons/deltas or otherwise?
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You can't completely avoid the epsilons and the deltas- but there is a quick proof as follows:
1) Show that if $f(x)=x$ then $lim_{x \rightarrow a}f(x)=a$
2) Show that if $f_1$ and $f_2$ are continuous functions, then so is $f_1 \cdot f_2$
3) Use induction.
voldemort
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Your point (1) is missing the assumption that $f$ is continuous. ā Jack M Apr 12 '14 at 17:10
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1@JackM: $f(x)=x$ is cont. Here one has to use epsilon delta argument :). ā voldemort Apr 12 '14 at 17:11
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Oh sorry, for some reason I interpeted $f(x)=x$ as $f(a)=a$. ā Jack M Apr 12 '14 at 17:17
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Let $x=a+h$. We then need to prove that $$\lim_{h \to 0} (a+h)^n = a^n$$ Let $n > 0$ and $h>0$ to begin with. We then have for sufficiently small $h$ $$a^n < (a+h)^n < a^n+2nh a^{n-1}$$ Now let $h \to 0$ and conclude using sandwich theorem. Repeat the same argument for $h < 0$. For $n < 0$, set $m = -n$ and conclude what you want.
user141421
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