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...
 A: 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.
A: There are a few common properties of limits that are quite useful to know. The one which is relevant here is the following:
If $f, g$ are continuous functions at the point $x_0$ and $g(x_0) \neq 0$, then $\frac{f}{g}$ is continuous at $x_0$ as well. This statement is the one you should prove using the epsilon-delta definition of a limit. Your specific example will follow.
A: Remember that the limit of a function $\lim_{x \to a} f(x) = L$ if and only if we can make
$$
\left|f(x) - L\right|
$$
small as $x \to a$. In our case, this is
$$
\left|\frac{3 + 7x}{5 + 5x} - \frac{3}{5}\right| = \left|\frac{5(3+7x) - 3(5 + 5x)}{5(5+5x)}\right| = \left|\frac{20x}{5(5+5x)}\right|
$$
So suppose now that you are given some $\varepsilon > 0$. Can you choose a $\delta > 0$ so that if $|x| < \delta$ you can make this term small?
A: The simplest method of proving $\lim_{x \to a} f(x) = L$ is to start from the inequality $|f(x)-L| < \epsilon$ and attempt to solve that inequality for the quantity $|x-a|$. If you can get your solution into the form 
$|x-a| < $(an expression involving only $\epsilon$ and constants and not $x$) 
then you set that expression on the right hand side equal to $\delta$ and you are done. Unfortunately that is rarely possible, but it is a good starting strategy and very often leads to a complete proof. In this case, for instance, starting from
$$(1) \qquad \biggl| \frac{3+7x}{5+5x} - \frac{3}{5} \biggr| < \epsilon
$$
and simplifying and solving you can show that inequality (1) is equivalent to  
$$(2) \qquad |x| < \frac{\epsilon}{4} |1+x|
$$ 
But the right hand side of (2) depends on $x$ so we are not yet done.
From this point you can observe that if $x < \frac{\epsilon}{4}$ and if $x < 1$ then inequality (2) is true and so inequality (1) is true. In other words, if $x < \delta = Min\{\epsilon/4,1\}$ then (1) is true. QED. 
