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

[MarkFL]

Here is the question:

Maths: Calculus > Functions?

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Please help with the question above. It was too hard to type out so I uploaded the image

Thanks

Here is a link to the question:

Maths: Caluclus > Functions? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello D,

We are given:

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

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

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

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

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

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

$$-1+a=8+b$$

$$a=9+b$$

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

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

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

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

$$2(0)^2+b=2^0+1$$

$$b=2\,\therefore\,a=11$$

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 http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.