I don't understand how to do this question. I've done a sketch, but wouldn't the limit as x tends to -2 from above be 2? I mean, it's on the actual point that it's tending too... However, I get the feeling that a is -3 from the question... But why? And how do I find b and c?
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$\begingroup$If limit exists everywhere then, in particular it exists at $-2$.
If $\lim_{x\to-2^+}f(x)=-3$ we must have $\lim_{x\to-2}f(x)=-3$. This implies that $$-3=\lim_{x\to-2^{-}}f(x)=\lim_{x\to-2^-}a=a,$$
where the second equality is because $f(x)=a$ for $x<-2$ close to $-2$.
To get $b$ check what happens at $x=3$.
At $x=3$ we need to compute the limit (not just compute the value of $f(3)$). Since the function is defined by pieces, it is likely to be simpler to compute lateral limits than just limit. Then let us look at
$$\lim_{x\to3+}f(x)$$ and $$\lim_{x\to3-}f(x)$$
We have that
$$\lim_{x\to3+}f(x)=\lim_{x\to3+}(2x+16)=22$$
and that $$\lim_{x\to3-}f(x)=\lim_{x\to3-}(x^2+bx+c)=9+3b+c$$ So, we must have $9+3b+c=22$. That is one equation relating $b$ and $c$.
But at $x=-2$ we didn't do the complete work. What about looking at $\lim_{x\to-2+}f(x)$?
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