# evaluating limits numerically.

• May 15th 2009, 04:26 PM
VonNemo19
evaluating limits numerically.
Fill in the following table to find $\lim_{x\to-2}f(x)$ where $f(x)=\frac{x^3+2x^2-x-2}{x^2-4}$. Give table values to four signifigant digits.

_x__-2.5__-2.1__-2.01__-2.002__-2.0001__-2.00001__-2
f(x)_____________________________________________?

_x__-1.5__-1.9__-1.99__-1.999__-1.9999__-1.99999__-2
f(x)_____________________________________________?

Based on this table my estimate, to two significant digits, for $\lim_{x\to-2}f(x)=$__?___

Anybody?

Hey guys. I can pretty much do the table, but I put it up there so that you guys would understand that we are evaluating the limit numerically, not algebraically. That's why we were asked to give an answer to with the given margin of error. Anybody got any ideas ?

Is this a bad question or something? usually I get a much quiker response.
• May 15th 2009, 05:32 PM
HallsofIvy
If you have already filled in the table it should be obvious what the answer is! If you don't feel like filling in all blanks, what did you get for x= -2.00001 and x= -1.99999?

By the way, there is no mention of "margin of errror" here. The "two significant figures" just asks you to reduce from the four significant figures ("thousands" place) to two ("tenths" place).
• May 15th 2009, 06:13 PM
VonNemo19
.75? Are you sure that the question is asking me to round this answer off to the tenths place? I thought that maybe the two significant digits were the ones showing 7 and 5. So the answer is .8? Unneccessary zeros count as significant digits?
• May 15th 2009, 06:26 PM
derfleurer
0.753486 to 2 sig figs is 0.75

0.758432 to 2 sig figs is 0.80

0.758432 to 1 sig fig is 0.80
• May 15th 2009, 06:45 PM
VonNemo19
Sooo,- 0.75 it is then?
• May 15th 2009, 06:55 PM
derfleurer
$-\frac{3}{4}$ it looks like.