Limits

J

jedwa1216

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Feb 5, 2011
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6
Hi,

I have been working on this problem for hours and cannot get a consistent answer, I either get 1/3 or -1/9. I have tried this problem in the Mathway calculator but really no help. I need to understand how to work this problem on my own. I am in a beginning calculus class so we are supposed to factor. I have been trying to get the negative exponents out of the numerator but I am missing a step somewhere. Thanks for any help you can provide. Jeff.

lim[x:0,((x+3)^-1-3^-1)/(x)]

is it correct to do this: [(1/x+3)-(1/3)]/x to get rid of the negative exponents?
 
Your problem is idfficult to read without grouping symbols.

I suppose it is:

\(\displaystyle \frac{(x+3)^{-1}-3^{-1}}{x}\)

If so, yes, you can write:

\(\displaystyle \frac{\frac{1}{x+3}-\frac{1}{3}}{x}=\frac{1}{x(x+3)}-\frac{1}{3x}=\frac{-1}{3(x+3)}\)

You should be able to see the limit at this point. It is one of your previous answers.
 
Thanks so much. Sorry the problem was kind of jammed together. I cut and pasted it from Mathway. I should have spread it out some. My problem was I was not distributing the x in the denominator of the original equation to both sides of the new denominators before doing the subtraction. Thanks again. Great board.
 
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