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

Staff member

MarkFL

Staff member
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.