So the idea I thought of doing was by induction.

$\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x-x_0| < \epsilon $

By the definition of the limit choose $\delta=\epsilon$ and the statement holds for $n=1$. Thus $\lim_{x \rightarrow x_0} = x_0 = f(x_0)$

Now here is the inductive step.

Next assume that $\lim_{x \rightarrow x_0} x^n$ exists. Another words, $\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x^n-x_0^n| < \epsilon $

$\forall\; \epsilon > 0\;\; \exists \;\delta > 0$ such that $|x-x_0| < \delta \implies |x^{n+1}-x_0^{n+1}| < \epsilon $

Let $\epsilon >0$ then $|x-x_0| < \delta \implies |x-x_0|^n|x+x_0| < \epsilon$

I am having trouble proceding from here any hints or advice would be greatly appreciated.


Induction isn't really necessary here. It is enough to just work with arbitrary $n$ to show the argument works for whatever $n$ you choose.

Indeed, fix $\varepsilon > 0$. We want $|x^{n+1} - x_0^{n+1}| < \varepsilon$. Notice that

$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n|$$

When $x$ is in an interval around $x_0$, say $x \in [x_0 - 1, x_0+1]$, the sum on the RHS has a bound, call this bound $M$.

Hence, taking $\delta = \min(\dfrac{\varepsilon}{M}, 1)$ we get

$$|x^{n+1} - x_0^{n+1}| = |x - x_0||x^n + x^{n-1}x_0 + \dots + x x_0^{n-1} + x_0^n| < \delta M = \varepsilon$$

so we are done.

  • $\begingroup$ How do you know that $|x^n+x^{n-1}+...|$ is contained in |$x_0 -1, x_0+1|$ $\endgroup$ – adam Jun 10 '16 at 4:12
  • $\begingroup$ Also did you assume $\delta \leq 1$? $\endgroup$ – adam Jun 10 '16 at 4:13
  • $\begingroup$ @adam I'm not saying the value of that is contained in the interval, I'm saying the value of that is bounded when you take $x$ is in that interval. Thus, to establish that bound I must take $\delta \leq 1$, which is fine because we only require the existence of such a $\delta$. $\endgroup$ – MCT Jun 10 '16 at 13:37

Hint: if $f$ is continuous and $g$ is continuous then $fg$ is continuous. This is not that hard to show (ask if you need help). Then apply induction.

  • $\begingroup$ That would be a lemma I would have to prove. The product of two continuous functions is continuous. I think he wanted the limit def. proof though. $\endgroup$ – adam Jun 10 '16 at 4:21
  • $\begingroup$ @adam It's not hard though! You can prove that lemma fairly easily using the limit definition. $\endgroup$ – user223391 Jun 10 '16 at 4:24
  • $\begingroup$ Assume $fg$ is not continuous, then $\exists c$ such that $\lim_{x\rightarrow c fg(x)} \neq fg(c)$ However by the def. of limit product, $fg= \lim_{x\rightarrow c} f(x) \lim_{x\rightarrow c} g(x)$. So at least one function $f$ or $g$ is not continuous at $c$ which contradicts that $f$ and $g$ are continuous. Would that be how it would go? $\endgroup$ – adam Jun 10 '16 at 4:31
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    $\begingroup$ @adam Your proof should involve some $\epsilon$s. :) $\endgroup$ – user223391 Jun 10 '16 at 4:33
  • $\begingroup$ At first I thought you meant the composition rather than the product of $f$ and $g$, and was going to comment that it doesn't work for primes $n$ :) $\endgroup$ – MCT Jun 10 '16 at 19:21

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