How to prove that $\lim_{x\rightarrow n^-}e^{\frac{n}{x-n}} = 0$ using a $\delta-\varepsilon$ proof.

Question

I'm trying to prove that $e^{\frac{n}{x-n}}\rightarrow 0$ as $x\rightarrow n^-$ where $n>0$ using a $\delta-\varepsilon$ proof. Here's what I've got thus far.

$\mid e^{\frac{n}{x-n}} \mid = e^{\frac{n}{x-n}} < \varepsilon \Leftrightarrow \frac{n}{x-n} < \ln(\varepsilon) \stackrel{x<n}{\Leftrightarrow} n > (x-n)\ln(\varepsilon) \Leftrightarrow -n<(n-x)\ln(\varepsilon) \stackrel{0<\varepsilon<1}{\Leftrightarrow} -\frac{n}{\ln(\varepsilon)}>n-x \stackrel{x<n}{=} \mid x-n \mid$. Thus if we let $0<\varepsilon<1$ then we can set $\delta = -\frac{n}{\ln(\varepsilon)}$. When $\varepsilon\geq1$ I don't know what to do. How do I find a $\delta$ for that case?

Answer 1

In any $\epsilon$-$\delta$ proof, it is not restrictive to consider only "small" values of $\epsilon$. In your case $0<\epsilon<1$ is good. If you prove the statement for any $\epsilon$ small, the same statement will trivially hold for $\epsilon$ "large", since $\epsilon < \epsilon+ h$ for every $h > 0$.

Answer 2

If you find a $\delta$ that shows $n-x < \delta \implies f(x) < \epsilon_1 < 1$. Then if $\epsilon_2 \ge 1$, the same $\delta$ will show $n-x < \delta \implies f(x) < \epsilon_1 < 1 \le \epsilon_2$.

It is perfectly valid to say, "we will restrict ourselves only to $\epsilon$s less than $c$" (It is NOT okay to say we will restrict ourselves to $\epsilon$s larger than $c$") or to say "since $\epsilon$ can be arbitrarily small we may assume $0 < \epsilon < 10^{-5,000}$". Those are FINE statements.

If you don't feel comfortable with that you can say. "If $\epsilon < 1$ this works. If $\epsilon \ge 1$ we can replace $\epsilon$ with $\epsilon_2$ so that $0 < \epsilon_2 < 1 \le \epsilon$."

Or as $\delta$ is a value dependent upon $\epsilon$ you think of it as a function. $\delta_{\epsilon} = - \frac n{\ln(\epsilon)}$. If we need the the input to be $\le 1$ but we dont know if $\epsilon$ is we can say $\delta_{\min(\epsilon, \frac 12)}$. That will prove if $n - x < \delta$ then BOTH $f(x) < \frac 12$ and $f(x) < \epsilon$. If $\epsilon \ge 1$ then $f(x) < \frac 12 < \epsilon$.

Perfectly valid.