Use the epsilon-delta definition of limits to evaluate the limit $\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ The Problem: Evaluate the given limit and prove your conclusion using only definitions from the epsilon-delta limit definition: $$\lim_{x \rightarrow1}\frac{x^2-x-2}{2x-3}$$
First I evaluated the limit and found the limit of the function to be 2.  For every $\varepsilon > 0$, there exists a $\delta$ so that $|x-1|< \delta \Rightarrow \frac{x^2-x-2}{2x-3}-2 < \varepsilon$.  After doing some algebra on the epsilon inequality, I end up with $|\frac{x^2-5x+4}{2x-3}| < \varepsilon$.  Since I can't prove the limit by defining x in terms of $\varepsilon$ and $x$, my professor has told my class that I should choose a specific value for delta for the purpose of substituting it into the inequality to find an x in terms of epsilon alone.  
I let $\delta=1$, and it follows from $|x-1|<1$ that $|x|<2$, and that $|\frac{1}{2x-3}|>1$ (for eliminating the denominator.  I think that $|\frac{1}{2x-3}|>1$ has set me back in solving this rigorously, since the expression is greater than, rather than less than 1. How do I proceed?
 A: Note that $\frac{x^2-x-2}{2x-3} -2  = \frac{1}{2}(1-\frac{5}{2x-3})(x-1)$, so we have $|\frac{x^2-x-2}{2x-3} -2 | = \frac{1}{2}|1-\frac{5}{2x-3}||x-1|$.
The $x \mapsto |1-\frac{5}{2x-3}|$ part looks like

We see that we need to 'stay 'away' from $x=\frac{3}{2}$. If we choose $1-\frac{1}{5} < x < 1+\frac{1}{5}$, we see that $3-2x > 3-2(1+\frac{1}{5}) = \frac{3}{5}$, and so 
$|1-\frac{5}{2x-3}| = |1+\frac{5}{3-2x}| < 1+ 5 (\frac{5}{3}) = \frac{28}{3}$.
Then we have 
$|\frac{x^2-x-2}{2x-3} -2 | \le  \frac{1}{2}\frac{28}{3}|x-1|$.
Now, in addition to $|x-1| < \frac{1}{5}$, we need another constraint on $|x-1|$ so that the last expression is less than $\epsilon$.
A: Let $f(x)$ be our function. It is clear that the limit will be $2$. Calculation shows that 
$$f(x)-2=\frac{x-4}{2x-3}(x-1).$$
As a preliminary step, insist that $\delta\lt \frac{1}{4}$. So $x$ is in the interval $(\frac{3}{4},\frac{5}{4})$. That ensures that $|2x-3|\gt \frac{1}{2}$.
Note that it also ensures that $|x-4|\lt 4$. It follows that if $|x-1|\lt \delta$ then 
$$|f(x)-2|\lt 8|x-1|.$$
Thus to make sure that $|f(x)-2|\lt \epsilon$, it is enough to take $\delta=\min(\frac{1}{4}, \frac{\epsilon}{8})$. 
