How Should I correctly attempt $\lim\limits_{x\to2}\frac{|x^2-4|}{x-2}$

Question

How should I evaluate this function's limits?

$$\lim_{x\to2}\frac{|x^2-4|}{x-2}$$

When graphing the function, I know its limit does not exist. My first attempt was to reconstruct the numerator as $(x+2)(x-2)$ and concludes its limits is $4$.

How should I evaluate its limit without graphing it, and what's the problem with my first attempt. Is there any general approach to deal with those problems?

Any help or comments would be appreciated!!!

Answer 1

The key is to understand the function $\frac{|x|}{x}$, which is equal to $1$ when $x> 0$ and equal to $-1$ when $x<0$. Similarly,

$$\frac{|x^2-4|}{x-2} = |x+2|\cdot \frac{|x-2|}{x-2} = \begin{cases} \hphantom{-}|x+2|,&x-2>0,\\ -|x+2|,&x-2<0. \end{cases}$$

Therefore, $$\lim_{x\to 2^+}\frac{|x^2-4|}{x-2} = \lim_{x\to 2^+}|x+2|=4,$$ $$\lim_{x\to 2^-}\frac{|x^2-4|}{x-2} = \lim_{x\to 2^-}(-|x+2|)=-4.$$

Since the limits from the left and the right are not equal, the limit doesn't exist.

Answer 2

\begin{align} \lim_{x \to 2^+}\frac{|(x+2)(x-2)|}{x-2} &= \lim_{x \to 2^+} \frac{|x+2|}{\text{sign}(x-2)}=4\\ \lim_{x \to 2^-}\frac{|(x+2)(x-2)|}{x-2} &= \lim_{x \to 2^-} \frac{|x+2|}{\text{sign}(x-2)}=-4 \end{align}

Hence, the limit does not exist.

Answer 3

When $x\rightarrow 2^{+}$, $x> 2\Rightarrow x-2>0\Rightarrow |x-2|= x-2$. Thus the limit of the given expression is +4. When $x\rightarrow 2^{-}$, $x<2\Rightarrow x-2<0\Rightarrow |x-2|=2-x$. Thus your limit now becomes -4.

Since left-hand limit is not equal to right-hand limit, we conclude limit doesn't exists.