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.

Thanks in advance!


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.

  • $\begingroup$ How is $1/2$ greater than $2/3$? $\endgroup$ – MathIsLife12 Oct 16 '18 at 4:53
  • 1
    $\begingroup$ thanks, i made a mistake. $\endgroup$ – Siong Thye Goh Oct 16 '18 at 6:50

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