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D's question at Yahoo! Answers regarding the existence of limits

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MarkFL

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Feb 24, 2012
13,775
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello D,

We are given:

\(\displaystyle f(x)=\begin{cases}
\cos\left(\frac{\pi x}{2} \right)+a && x<-2 \\
100 && x=-2 \\
2x^2+b && -2<x<0 \\
2^x+1 && 0<x \\
\end{cases}
\)

In order for \(\displaystyle \lim_{x\to-2}f(x)\) to exist, we require:

\(\displaystyle \lim_{x\to-2^{-}}f(x)=\lim_{x\to-2^{+}}f(x)\)

Now, using the definition of $f(x)$, we find this means:

\(\displaystyle \lim_{x\to-2^{-}}\left(\cos\left(\frac{\pi x}{2} \right)+a \right)=\lim_{x\to-2^{+}}\left(2x^2+b \right)
\)

\(\displaystyle \cos\left(\frac{\pi\cdot2}{2} \right)+a=2(2)^2+b\)

\(\displaystyle -1+a=8+b\)

\(\displaystyle a=9+b\)

In order for \(\displaystyle \lim_{x\to0}f(x)\) to exist, we require:

\(\displaystyle \lim_{x\to0^{-}}f(x)=\lim_{x\to0^{+}}f(x)\)

Now, using the definition of $f(x)$, we find this means:

\(\displaystyle \lim_{x\to0^{-}}\left(2x^2+b \right)=\lim_{x\to0^{+}}\left(2^x+1 \right)\)

\(\displaystyle 2(0)^2+b=2^0+1\)

\(\displaystyle b=2\,\therefore\,a=11\)

This ensures the limits exist, and while there is a discontinuity at $x=-2$, this is allowed as the function need not have the value of the limits at that point.

To D and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.