show limit by definition

Question

I need to prove $\displaystyle \lim _{x\to 2-} \left(\frac{|x-2|}{x^2-4}\right)=\frac{-1}{4}$

I know the definition $\forall \varepsilon >0, \exists \delta >0, 0>2-x>\delta$ then $\left|\left(\dfrac{|x-2|}{x^2-4}\right)+\dfrac{1}{4}\right|<\varepsilon$

And I also know how to calculate a limit but I don't know how to prove that a limit is correct

Answer 1

Guessing a value for $\delta$: Since we're interested in the behavior of $|x-2|/(x^2-4)$ as $x$ tends to $2$ from its left, we can assume that $x<2$. Then $x-2<0$, so $|x-2|=-(x-2)$. Using the identity

$$x^2-4=(x-2)(x+2)$$

we deduce that

\begin{align*} \left|\frac{|x-2|}{x^2-4}+\frac{1}{4}\right|<\varepsilon &\iff \left|\frac{-(x-2)}{(x-2)(x+2)}+\frac{1}{4}\right|<\varepsilon\\ &\iff \left|-\frac{1}{x+2}+\frac{1}{4}\right|<\varepsilon\\ &\iff \left|\frac{1}{x+2}-\frac{1}{4}\right|<\varepsilon\\ &\iff -\varepsilon<\frac{1}{x+2}-\frac{1}{4}<\varepsilon\\ &\iff \frac{1}{4}-\varepsilon<\frac{1}{x+2}<\frac{1}{4}+\varepsilon\\ &\iff \frac{1}{\frac{1}{4}+\varepsilon}<x+2<\frac{1}{\frac{1}{4}+\varepsilon}\\ &\iff \frac{1}{\frac{1}{4}+\varepsilon}-4<x-2<\frac{1}{\frac{1}{4}+\varepsilon}-4\\ \end{align*}

We've assumed that $x-2<0$, so

$$\frac{1}{\frac{1}{4}+\varepsilon}-4<x-2<0$$

Thus, choosing $\delta=-\left(\frac{1}{1/4+\varepsilon}-4\right)$ and working backwards, we will end up with

$$\left|\frac{|x-2|}{x^2-4}+\frac{1}{4}\right|<\varepsilon$$

so

$$\lim_{x\to 2^-}\frac{|x-2|}{x^2-4}=-\frac{1}{4}$$

Answer 2

Given $\varepsilon>0$, we want to find $\delta>0$ such that if $0<2-x<\delta$, then $$\left|\frac{|x-2|}{x^2-4}+\frac{1}{4}\right|<\varepsilon $$

We know that $|x-2|=2-x$ since $x<2$. Suppose for a moment that we have such a $\delta>0$ which satisfies $0<2-x<\delta$. We also have $x^2-4=(x+2)(x-2)=-(x+2)(2-x)$ so $$\left|\frac{|x-2|}{x^2-4}+\frac{1}{4}\right|=\left|\frac{1}{x+2}-\frac{1}{4}\right|=\left|\frac{2-x}{4(x+2)}\right| =\frac{2-x}{4|x+2|}<\frac{\delta}{4|x+2|}$$

We now need a lower bound for $x+2$. We have $$-\delta<x-2<0$$ so $$4-\delta<x+2<4$$ Assuming $\delta\leq 1$ means $3\leq 4-\delta<x+2$ so $$0<\frac{1}{x+2}\leq\frac{1}{3}$$ If we take $\delta=\min\{1,\varepsilon\}$ for any $\varepsilon>0$, then we will have $$\frac{\delta}{4|x+2|}=\frac{\delta}{4(x+2)}\leq\frac{\delta}{4\cdot 3}\leq\frac{\varepsilon}{12}<\varepsilon $$