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akotoh · Jul 13, 2009, 05:39 AM

can anyone check if my answer is correct or not. The topic is about limits.
Advance thank you! :)

1. lim 8/(t-4)
t->4^+
my answer: positive infinity

2. lim (s-5)/(3s-2)
s->(2/3)^+
my answer: negative infinity

3. lim (1/t)-(1/t^2)
t->0^-
my answer: positive infinity

4. lim 2x/(x^2 - 2x)
x->2^-
my answer: positive infinity

5. lim (5-r)/((r-5)^2)
r->5^+
my answer: positive infinity

6. lim (x/x+2)+(1/x^2-4)
t-> -2^+
my answer: negative infinity

please correct my answer if it is wrong.

galactus · Jul 13, 2009, 06:02 AM

Originally Posted by akotoh

can anyone check if my answer is correct or not. The topic is about limits.
Advance thank you! :)

1. $\lim_{t\to 4^{+}}\frac{8}{t-4}$

my answer: positive infinity

Correct.

2. $\lim_{s\to \frac{2^{+}}{3}}\frac{s-5}{3s-2}$

my answer: negative infinity

Correct.

3. $\lim_{t\to 0^{-}}\frac{1}{t}-\frac{1}{t^{2}}$

my answer: positive infinity

should be ${-\infty}$

4. $\lim_{x\to 2^{-}}\frac{2x}{x^{2} - 2x}$

my answer: positive infinity

Should be ${-\infty}$

5. $\lim_{r\to 5^{+}}\frac{5-r}{(r-5)^{2}}$

my answer: positive infinity

Should be ${-\infty}$

6. $\lim_{t\to -2^{+}}\frac{x}{x+2}+\frac{1}{x^{2}-4}$

my answer: negative infinity

Correct. To help see why, look at their graphs.

akotoh · Jul 13, 2009, 06:07 AM

Thank you! :)

ebaines · Jul 13, 2009, 06:09 AM

I will assume that your notation :

lim
t -> 4^+

means the limit as t approaches 4 from the positive side. With this understanding - your first two answers are correct, but the last four are not. There are a couple of ways to check these:
1. You can graph the functions and see how they behave.
2. A quick check is to plug in a value for the variable that is very close to the limit, but offset a bit in the direction of interest. For example, in the third problem you have

lim (1/t)-(1/t^2)
t->0^-

So try plugging in t = -.01, and you get:
1/(-.01) - 1/(-.01)^2 = -100 - 10000 = -10100. It seems clear that this function is heading off to negative infinity as t approaches 0 from the negative side.

As for problem 6 - please clarify that what you meant is this:

$<br /> \lim _ {t \to -2^+} (\frac x {x+2})+ (\frac 1 {x^2-4})<br />$

akotoh · Jul 13, 2009, 06:11 AM

wait! In number 5. I just noticed.
I wrote r->5^+ and you wrote r->5^- .
Is the answer in number 5 is still negative infinity? :)

galactus · Jul 13, 2009, 06:14 AM

That was a typo on my part. It should be negative infinity if it is approaching 5 from the right.

akotoh · Jul 13, 2009, 06:18 AM

Ok! Thanks a lot! :)

akotoh · Jul 13, 2009, 07:07 AM

Is my answer in #6 correct or not?
I think my answer is wrong. :( it should be positive infinity?

ebaines · Jul 13, 2009, 07:26 AM

Number 6 is negatve infinity.