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What inequalities is used to show that $f(x)=c^x$ is continuous for some constant $c>0$ I am trying to use $\epsilon-\delta$ definition of continuity. That is

$f$ is continuos at $x_0$ if for every $\epsilon>0$ there exists a $\delta>0$ such that $|x-x_0| < \delta$ implies $|f(x)-f(x_0)|< \epsilon$.

So, basically how to do make sure of the following \begin{align} |c^x-c^{x_0}| \le a|x-x_0|=a\delta, \text{ where $a$ is some constant} \end{align} and let $\delta \le \epsilon/a$.

My question what inequality is used to get the part $|c^x-c^{x_0}| \le a|x-x_0|$.

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  • $\begingroup$ What is your recent definition of $c^x$? It could be $e^{x·\ln c}$, then you only have to use the continuity of the exponential function. Which then begs the question of what your latest definition of $\exp$ was. $\endgroup$ – Lutz Lehmann Oct 29 '15 at 22:12
  • $\begingroup$ I see. Thanks for clarifying. Lets assume that exponential is defined in terms of infitint series (taylor series) $\endgroup$ – Boby Oct 29 '15 at 22:33
  • $\begingroup$ Then use that $|e^{x+h}-e^x|=e^x·|e^h-1|\le e^x·\frac{|h|}{1-\frac{|h|}2}$. $\endgroup$ – Lutz Lehmann Oct 29 '15 at 22:55
  • $\begingroup$ Thanks. This is exactly what I wanted. Is there a proof of this inequality somewhere or does it have a name? $\endgroup$ – Boby Oct 29 '15 at 23:09
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    $\begingroup$ This is a a standard bound. You estimate the remainder of the exponential series by a geometric series, $$|e^h-1-h-…-h^{n-1}/(n-1)!|\le |h|^n/n!·(1+|h|/(n+1)+|h|^2/(n+1)^2+…)\\=|h|^n/n!·1/(1-|h|/(n+1)).$$ $\endgroup$ – Lutz Lehmann Oct 29 '15 at 23:26
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HINT You won't be able to find a single $a$ applicable for all values of $x$ and $\epsilon$ that proves continuity by taking $\delta \leq \epsilon /a$. However, the definition does not demand that degree of uniformity: $\delta$ is allowed to depend on both $\epsilon$ and $x$.

The problem is much easier if you allow yourself that freedom.

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  • $\begingroup$ I think I understand that. However, I still need to figure out how to bound $|c^x-c^{x_0}| \le a(x,c)|x-x_0|$, right? Is there such an inequality? $\endgroup$ – Boby Oct 29 '15 at 22:56

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