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$$\lim \limits_{h \to 0} \frac{\frac{1}{5+h}-\frac{1}{5}}{h}$$

I simplified, even made a table and came up with $-1/25$ but it appears to be incorrect, according to my submit module.

If anyone has any tips or tricks on finding these limits, it would be greatly appreciated! I understand that the one with h and x will have to be in terms of x in some form or another.

$$\lim \limits_{h \to 0} \frac{\frac{1}{x+h+3}-\frac{1}{x+3}}{h}$$

EDIT: What a typo, it is supposed to be subtraction on the top.

I'm going to add the full original problem to make sure I didn't make any mistakes early on, so I know if the module has a wrong answer or not.

The problem states:

$$f(x)=\frac{1}{x+3}$$

Compute the limit of the difference quotients $$\lim \limits_{h \to 0} \frac{f(2+h)-f(2)}{h}$$

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Sorry, huge mistake, I meant to put -, not +. –  BKaylor Sep 4 '11 at 0:48
    
Your first answer is correct. (You can mentally check that since it is also the derivative of $1/x$ at $x=5$.) –  Srivatsan Sep 4 '11 at 0:51
    
I don't know what a "submit module" is. If you are saying you typed -1/25 into some software, and you were told it was wrong, then it could be the submit module was having a senior moment, or it could be a syntax thing. Maybe you have to type in -(1/25), or -0.04, or - 1 / 25, or some other variation on -1/25. –  Gerry Myerson Sep 4 '11 at 6:45

2 Answers 2

up vote 3 down vote accepted

It looks to me as if you correctly calculated $$\begin{align*} \lim\limits_{h\to 0}\frac{\frac{1}{5+h}-\frac{1}{5}}{h} &= \lim\limits_{h\to 0}\frac{5-(5+h)}{5h(5+h)}\\ &= \lim\limits_{h\to 0}\frac{-1}{5(5+h)}\\ &= \frac{-1}{25}. \end{align*}$$ This is the problem that I’d have expected to see if you’re doing difference quotients, but if the problem was really to calculate $$\lim\limits_{h\to 0}\frac{\frac{1}{5+h}+\frac{1}{5}}{h},$$ as you’ve written it, then what you’ve done is indeed wrong: in that case the numerator approaches $\frac25$, so the limit is $\infty$ as $h \to 0^+$ and $-\infty$ as $h \to 0^-$.

Before I say anything about the second problem, then, let me ask: are you sure that you’ve copied it correctly? I’d expect a minus sign in the numerator.

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You're right, I did copy it down incorrectly. I corrected the original post. –  BKaylor Sep 4 '11 at 2:43
    
@BKaylor: Okay. Then it appears that the answer in the module is wrong. If you do exactly the same thing with the other problem, you should be fine: the numerator should be $-h$ after you simplify, so you’ll be able to cancel factors of $h$ and be left with a limit that evaluates to $\frac{-1}{(x+3)^2}$. –  Brian M. Scott Sep 4 '11 at 4:18

From the way the problem is stated, I think you're supposed to do it by recognizing that the limit equals the derivative $f'(2)$.

(Or else, if this problem was given before talking about derivatives, it's a warm-up problem to get a feel for the kind of limits you will be doing when you get to derivatives...)

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