# $\epsilon - \delta$ proof of the limit $\lim_{x\to1}\frac{1}{x-\frac{3}{2}}=-2$.

I want to use the $$\epsilon -\delta \$$definition of a limit to prove that

$$\lim_{x\rightarrow 1}\frac{1}{x-\frac{3}{2}}=-2.$$

My attempt:

$$\left | f(x)-L \right |=\left | \frac{1}{x-\frac{3}{2}} + 2 \right |= 4\left | \frac{x-1}{2x-3} \right |<\epsilon.$$ This implies $$\left | x-1 \right |<\frac{\epsilon \left | 2x-3 \right |}{4}.$$ Since $$\delta$$ can only be in terms of $$\epsilon$$, we need to somehow change the $$\left | 2x-3 \right |.$$ We know that $$0<\left | x-1 \right |<\delta.$$ Let's bound $$\delta$$ so that $$\delta \leq 1.$$ Then, $$-3<\left | 2x-3 \right |<1.$$ This is where I am a little confused. Obviously $$\left | 2x-3 \right |$$ is always greater than $$-3$$ since it positive. What value then do I plug in for $$\left | 2x-3 \right |$$? Once I find this, I can complete the proof by setting $$\delta=\min\left \{ 1, \text{missing value} \right \}$$ and then doing some algebra. Clearly this is not a fully written out proof- I just wrote what was needed to explain my question.

If $$\delta < \frac14$$, then $$|x-1| < \delta \implies \frac34 < x < \frac54 \implies \frac32<2x<\frac52 \implies -\frac32 < 2x-3 < -\frac12.$$
Hence we have $$\frac12 < |2x-3| < \frac32$$
$$\frac23 < \frac1{|2x-3|} < 2$$
Hence $$|f(x)-L|=\frac{4|x-1|}{|2x-3|}<8\delta$$
Hopefully you can pick your $$\delta$$ now.
• How is $1/2$ greater than $2/3$? – MathIsLife12 Oct 16 '18 at 4:53