# Issues Calculating Limits Using Derivatives

I'm pretty bad at math, and Im doing some practice questions for school, I was wondering if someone could help me with this question, I cant get the right answer no matter what I try. Here's the question:

$$f(x) = \left\{\begin{array}{ll} -\frac{3}{x+3},&\text{if } x \lt -3;\\ 2x+9, &\text{if }x \gt -3. \end{array}\right.$$

Calculate the following limits:

a) $\displaystyle \lim_{x\to -3^-} f(x)$

b) $\displaystyle \lim_{x\to -3} f(x)$

c) $\displaystyle \lim_{x\to -3^+} f(x)$

I've gotten (These are right)

B) DNE

C) 3

But I cant figure out A do I need the

$$\frac{f(x + h) - f(x)}{h}$$ or $$\frac{f(x)-f(a)}{x-a}$$ for that one? how would I work it out?

Here's my work if you want to look at what i did wrong:

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Why do you need the derivative? BTW, your b & c are mixed up –  Dennis Gulko Oct 5 '11 at 20:05
For (a), you have the form -3 ÷ "-0", where -0 is just a way of saying something that approaches zero from the left (and is hence negative). So this limit diverges to + infinity. –  The Chaz 2.0 Oct 5 '11 at 20:06
Please upload images using the standard host of this site, because then they will be guaranteed to be persistent. To do this, use the image link above the editing field or hit ctrl-G. –  t.b. Oct 5 '11 at 22:09

In what you write, you are not being asked to do a derivative, you are just being asked to do a limit. So you do not need to use either the Fermat or the Difference quotient limits. You just need to do the limit of the function.

As $x\to -3^-$, $f(x)$ will be evaluated using the formula $\displaystyle-\frac{3}{x+3}$. So $$\lim_{x\to -3^-}f(x) = \lim_{x\to -3^-}-\frac{3}{x+3}.$$

When $x\to -3^+$, $f(x)$ is evaluated using the formula $2x+9$, so you would do $$\lim_{x\to -3^+}f(x) = \lim_{x\to -3^+}(2x+9).$$

These limits should be done directly; neither of the derivative limits comes into play at all.

If you are being asked for the limits of the derivative of $f(x)$, i.e., $\displaystyle\lim_{x\to -3^-}f'(x)$, $\displaystyle\lim_{x\to -3^+}f'(x)$, $\displaystyle \lim_{x\to -3}f'(x)$, then first you should figure out the derivative for each of the two "parts" of $f(x)$ using whichever quotient you want, then take the limit.

For example, if $a\gt -3$, then $$f'(a) = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{x\to a}\frac{(2x+9)-(2a+9)}{x-a};$$ or, just as valid, $$f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} = \lim_{h\to 0}\frac{2(a+h)+9 - (2a+9)}{h}.$$ Then, whatever the answer is, you use this formula for $\displaystyle \lim_{x\to -3^+}f'(x)$. Similarly with the other limit.

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Alright that makes it a lot clearer, thank you –  Latency Oct 5 '11 at 20:58

When stuck with limits, sometimes drawing a graph could help.

as x approaches -3 from the negative side, the function -(3/(x+3)) approaches very large values of y. As x approaches -3 from the right, the function approaches 2*-3+9=3. As a result the function has no limit at x=-3 and is not continuous at this point.

In an exam, you probably can't relay on the plot, but the picture together with algebraic derivation sometimes help.

Wolfram provide a free on-line plotter. Also, you may want to check this: GraphPlotter

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