# Distinguishing and solving limits

Given: $f(x)=\left\{ \begin{array}{lcl} -3x+8 &\mbox{ if }&x\geq 3\\ \frac{4}{2 -x} &\mbox{ if }&x<3\\ \end{array} \right.$

Find the limit if it exists. If the limit does not exist, distinguish between that goes infinity or (-) infinity and DNE. If f(x) continuous at x=3 and x=-1? Why?

Little help here? I dont understand the steps to completing this question and how to approach it.

1. 2.
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For $x<3$, do you mean $\frac{4}{2-x}$? – Ian Coley Sep 18 '13 at 23:09
Yes not sure how to edit that – jayson Sep 18 '13 at 23:11

Hint: $f$ is continuous at $a$ if and only if $$\lim_{h\to 0}f(a+h)=f(a)=\lim_{h\to 0}f(a-h).$$

Additionally, $f$ is going to go bad at $x=2$, because $\frac{4}{2-2}=\frac{4}{0}$ is not good. Clearly $f$ is not continuous at $2$. However, what are the limits as calculated above?

Edit: you can you use $$\lim_{x\to a^-}f(x)=\lim_{h\to0}f(a-h)\;\text{ and }\;\lim_{x\to a^+}f(x)=\lim_{h\to 0}f(a+h).$$

We're going to find $\lim_{x\to 3^+}f(x)$. We see that $$f(3+h)=-3(3+h)+8=-1+3h,$$ so as $h\to0$, $f(3+h)\to -1$.

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Oh shoot I didnt put in the rest of the question. – jayson Sep 18 '13 at 23:16
Sorry about that – jayson Sep 18 '13 at 23:19
Not sure what your doing, mind showing me the steps and how to do for just one of the two questions? Thanks – jayson Sep 18 '13 at 23:28
Sure, one second. – Ian Coley Sep 18 '13 at 23:28