# limit and absolute absolute value problem

$$\lim_{x \to -2} \frac{2-|x|}{2+x}$$

If I calculate the left and right-hand limit I get different results.

Left hand side: $$\lim_{x \to -2^-}\frac{2+x}{2+x}=1$$

Right hand side: $$\lim_{x \to -2^+} \frac{2-x}{2+x}=\text{undefined}$$

My question is that my procedure is right or wrong?

• I get a solution from an unauthorized source what explains that it is 1 – Jobiar Hossain Feb 9 at 9:49
• Note that the modulus function is not continuous at $0$, but it is around a neighborhood of $-2$, hence you could compute the limit simply assuming $|x|=-x$. – Mefitico Feb 9 at 20:29

Since $$x \to -2$$, we can assume that $$x < 0$$ so that $$|x| = -x$$.

Then $$\frac{2-|x|}{2+x} = \frac{2+x}{2+x} = 1 \xrightarrow{x \to -2} 1$$ so the limit exists and it is equal to $$1$$.

Multiply and divide by $$2+|x|$$ and cancel: $$\lim_{x \to -2} \frac{2-|x|}{2+x}=\lim_{x \to -2} \frac{4-x^2}{(2+x)(2+|x|)}=\lim_{x \to -2} \frac{2-x}{2+|x|}=\frac{2-(-2)}{2+2}=1.$$

When $$x$$ is near $$-2$$ approaching it from the left, $$2+x$$ is equivalent to $$2-|x|$$ (if $$x<0$$, then $$x=-|x|$$):

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

When $$x$$ is near $$-2$$ approaching it from the right, $$2+x$$ also seems to be equivalent to $$2-|x|$$:

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

Since both one-sided limits are equal to the same number, the limit exists and is equal to $$1$$:

$$\lim_{x\to-2}\frac{2-|x|}{2+x}=1.$$

Even though the function itself is undefined at $$x=-2$$ because the denominator at that point is zero ($$f(-2)=\frac{2-|-2|}{2-2}=\frac{0}{0}$$), the limit of this function at $$x=-2$$ does exist and is equal to $$1$$.

• When x is near −2 approaching it from the right, 2+x also seems to be equivalent to 2−|x| ...sorry, I don't get it – Jobiar Hossain Feb 9 at 10:35
• I get it. you are amazing to explain – Jobiar Hossain Feb 9 at 10:54

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

• it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0 – Jobiar Hossain Feb 9 at 9:58
• @Jobiar Hossain Since $x\rightarrow-2$, we can assume that $x<0$ because for $x\geq0$ it's not interesting. – Michael Rozenberg Feb 9 at 9:59
• yes, but why do we not consider left hand side and right hand side limit ? I know this idea but it don't give information about left hand side limit and right hand side limit. – Jobiar Hossain Feb 9 at 10:02
• @Jobiar Hossain $x\rightarrow-2$. Id est, it's interesting what happens around $-2$, id est, for $x<0$. If $x\rightarrow-2$ so can be $x>-2$ and can be $x<-2$ by the definition of the limits. We don't need to consider two cases here. – Michael Rozenberg Feb 9 at 10:08
• If you draw the graph, you will find that there is a vertical Asymptote at x=-2 – Jobiar Hossain Feb 9 at 10:18

$$\frac{2-|-2|}{2+-2}=\frac{0}{0}$$
$$\frac{d}{dx}(2-|x|)=-\frac{x}{|x|}, \frac{d}{dx}(2+x)=1$$
evaluate $$\frac{-\frac{x}{|x|}}{1}$$ at $$x=-2$$ yields
$$\frac{-\frac{-2}{|-2|}}{1} = \frac{-(-1)}{1}=1$$