How do you show the limit of a function? Using the epsilon-delta definition of a limit, how would you show that $\lim_{x \to 0} \frac{3+7x}{5+5x} = \frac{3}{5}$? The lecturer's definition and solution are nothing to shout about...

By definition, for a given $\epsilon > 0$, you want to $\delta > 0$ such that if $|x| < \delta$, then
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| < \epsilon $$
We want to find a bound for $\left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right|$. Lets see
$$ \left| \frac{ 3 + 7x}{5+5x} - \frac{3}{5} \right| = \left| \frac{ 3 + 7x - 3x -3}{5(x+1)} \right| = \left| \frac{3x}{5(x+1)} \right| = \frac{ 3|x|}{5|x+1|} $$
Notice we want to control the $|x+1|$ term in the denominator. To do so, let us assume a priori that $\delta < \frac{1}{2} $. Then
$$ |x| < \frac{1}{2} \iff - \frac{1}{2} < x < \frac{1}{2} \iff \frac{1}{2} < x+1 < \frac{3}{2} \iff 2 > \frac{1}{x+1} > \frac{2}{3}$$
Therefore,
$$ \frac{ 3|x|}{5|x+1|} < \frac{ 2\cdot3 |x|}{5} = \frac{6}{5} |x|$$
and the last quantity is less than $\epsilon$ precisely if we choose $\delta = \frac{5}{6} \epsilon $. Since we assumed a priori that $\delta < \frac{1}{2}$, then the choice
$$ \delta = \min\left\{ \frac{1}{2}, \frac{5}{6} \epsilon \right\} $$
will work, and the proof is complete.
