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

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?

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  • $\begingroup$ I get a solution from an unauthorized source what explains that it is 1 $\endgroup$ – Jobiar Hossain Feb 9 at 9:49
  • $\begingroup$ 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 $|x|=-x$. $\endgroup$ – Mefitico Feb 9 at 20:29
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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 $1$.

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

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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 $1$.

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  • $\begingroup$ 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 $\endgroup$ – Jobiar Hossain Feb 9 at 10:35
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    $\begingroup$ I get it. you are amazing to explain $\endgroup$ – Jobiar Hossain Feb 9 at 10:54
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$$\lim_{x\rightarrow-2}\frac{2-|x|}{2+x}=\lim_{x\rightarrow-2}\frac{2+x}{2+x}=1.$$

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  • $\begingroup$ it occurs when you consider right hand side limit I mean x if x>0 and Left hand side limit -x if x<0 $\endgroup$ – Jobiar Hossain Feb 9 at 9:58
  • $\begingroup$ @Jobiar Hossain Since $x\rightarrow-2$, we can assume that $x<0$ because for $x\geq0$ it's not interesting. $\endgroup$ – Michael Rozenberg Feb 9 at 9:59
  • $\begingroup$ 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. $\endgroup$ – Jobiar Hossain Feb 9 at 10:02
  • $\begingroup$ @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. $\endgroup$ – Michael Rozenberg Feb 9 at 10:08
  • $\begingroup$ If you draw the graph, you will find that there is a vertical Asymptote at x=-2 $\endgroup$ – Jobiar Hossain Feb 9 at 10:18
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What about L'Hospital's Rule?

Since

$$\frac{2-|-2|}{2+-2}=\frac{0}{0}$$

then

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

evaluate $\frac{-\frac{x}{|x|}}{1}$ at $x=-2$ yields

$$\frac{-\frac{-2}{|-2|}}{1} = \frac{-(-1)}{1}=1$$

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