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Let $f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$. Find $\lim_{x\rightarrow -1}f(x)$.

I broke up each absolute value into parts:

$|1-x^2| = \begin{cases} 1-x^2 & -1 \leq x\leq 1 \\ -(1-x^2) & x>1,x<-1 \end{cases} $ , $|x+1| = \begin{cases} x+1 & x\geq -1 \\ -(x+1) & x<-1 \end{cases} $, $|x^2+x| = \begin{cases} x^2+x & x\geq0,x\leq -1 \\ -(x+1) & -1<x<0 \end{cases} $

Thus, when $x\rightarrow -1$, the function will approach the positive value, because by my definitions of the absolute values,each has the positive value at $x=-1$. So then you can take $\lim_{x\rightarrow -1}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1}\frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(x^2+x)}=\lim_{x\rightarrow -1}\frac{(4x-2)(x+1)}{(x+1)(-x+2)}=-2$.

But looking at the graph, the limit is -6. So I must have messed up in my absolute value declarations, most likely in the third one.

So, how would I correctly declare and solve this limit?

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    $\begingroup$ It seems that the factor $|x+1|$ in the denominator is causing trouble. Can you eliminate that by simplifying the expression for $f$? $\endgroup$ – player3236 Sep 27 at 14:55
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Note that\begin{align}\require{cancel}f(x)&=\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}\\&=\frac{|x+1|^{\cancel2}-3\cancel{|x+1|}|x-1|}{\cancel{|x+1|}(2-|x|)}\\&=\frac{|x+1|-3|x-1|}{2-|x|}\\&\to_{x\to-1}\frac{-6}1\\&=-6.\end{align}

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As an alternative, we have that by $x=y-1$ with $y\to 0$

$$\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\frac{y^2-3|2y-y^2|}{2|y|-|y(y-1)|}=\frac{|y|-3|2-y|}{2-|y-1|}\to \frac{0-6}{2-1}=-6$$

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$f(x) = \frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}$ $\lim_{x\rightarrow -1^-}f(x)=\lim_{x\rightarrow -1^-}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1^-} \frac{x^2+2x-3(x^2-1)+1}{2(-(x+1))-(x^2+x)}=\lim_{x\rightarrow -1^-}f(x)\frac{-2x^2+2x+4}{-x^2-3x-2}=\lim_{x\rightarrow -1^-} \frac{-2(x+1)(x-2)}{-(x+1)(x+2)}=\lim_{x\rightarrow -1^-} \frac{-2(x-2)}{-(x+2)}=-6$.

$\lim_{x\rightarrow -1^+}f(x)=\lim_{x\rightarrow -1^+}\frac{x^2+2x-3|1-x^2|+1}{2|x+1|-|x^2+x|}=\lim_{x\rightarrow -1^+} \frac{x^2+2x-3(1-x^2)+1}{2(x+1)-(-(x^2+x))}=\lim_{x\rightarrow -1^+}f(x)\frac{4x^2+2x-2}{x^2+3x+2}=\lim_{x\rightarrow -1^+} \frac{2(2x-1)(x+1)}{(x+1)(x+2)}=\lim_{x\rightarrow -1^+} \frac{2(2x-1)}{(x+2)}=-6$.

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