Continuity of $e^{-x^2/4k}$ I want to show that $f(x)=e^{-x^2/4k}$ (where $k>0$ is fixed) is continuous using an $\epsilon$,
$\delta$ argument. I've been trying to choose $\delta$ using $\ln$ somehow and I've also been trying to write $|f(x)-f(y)|$ in a form with $|x-y|$ in the exponent so that I can complete the proof, but I have gotten stuck. 

We have
$$|f(y)-f(x)| = |e^{-x^2/4k}||e^{-(y^2-x^2)/4k}-1|\leq |e^{-(y^2-x^2)/4k}-1|.$$
It is easy to show that $e^{-1/n} \rightarrow 1$ as $n \rightarrow \infty$. Hence, for every $\epsilon >0$ there exists $N \in \mathbb{N}$ such that if $n \geq N$,
$$|e^{-1/n}-1| = 1-e^{-1/n} < \epsilon.$$
Since $e^z$ is monotone increasing, if $|z| < 1/N$ then 
$$0 \leq1-e^{-|z|} <  1-e^{-1/N}< \epsilon$$
Since $x \mapsto x^2$ is continuous ,there exists $\delta > 0$ such that if $|y-x| < \delta$ then $|y^2-x^2|/4k < 1/N$.
In particular, take
$$\delta = \min\left(1, \frac{4k}{N[2|x|+1]}\right).$$
Then, if $|y-x| < \delta$,
$$|x|+|y| < |x| + |y-x +x|< 2|x| +  |y-x| < 2|x| +1,$$
and
$$|y^2-x^2| = |y-x||y+x| < |y-x|(|x|+|y|)< \delta [2|x|+1]< \frac{4k}{N}.$$
Assume WLOG that $x$ is fixed and $y > x$. 
Then, if $|y-x| < \delta$ we have, 
$$|f(y)-f(x)|  \leq |e^{-(y^2-x^2)/4k}-1|= 1 - e^{-|y^2-x^2|/4k}< \epsilon.$$
