# Calculus - Infinite Limits and Asymptotes (Several Problems)

• Oct 7th 2007, 08:02 PM
FalconPUNCH!
Calculus - Infinite Limits and Asymptotes (Several Problems)
Hi I just need help with several problems. I worked them out and I got an answer but I'm not sure if I did it correctly. Please check my answers.

Problem 1
Find the Limit:

Lim as X -> Positive Infinity of T^2 + 2/ T^3 + ^2 - 1

I don't want to put all my work because I don't know the shorthand to doing it and I'll just put the answer I got. I got 0 as the limit to this problem.

Problem 2
Find the Horizontal and Vertical Asymptotes of each curve:

Y = 1 + X^4/X^2-X^4

I got Horizontal = -1 and Vertical = 1, -1

Problem 3
Same as above

Y = 2e^x/e^x-5

I got Horizontal = 0 and Vertical = -1

Problem 4
Find the limit

Lim x-> positive infinity COSX

I got Lim doesn't exist because COSX is a periodic.
• Oct 7th 2007, 08:58 PM
FalconPUNCH!
Okay I changed some of my answers. I think they're all right but I just need someone to double check for me.
• Oct 7th 2007, 09:07 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
Hi I just need help with several problems. I worked them out and I got an answer but I'm not sure if I did it correctly. Please check my answers.

Problem 1
Find the Limit:

Lim as X -> Positive Infinity of T^2 + 2/ T^3 + ^2 - 1

I don't want to put all my work because I don't know the shorthand to doing it and I'll just put the answer I got. I got 0 as the limit to this problem.

did you mean $\displaystyle \lim_{T \to \infty} \frac {T^2 + 2}{T^3 + T^2 - 1}$? if so, you are correct

Quote:

Problem 2
Find the Horizontal and Vertical Asymptotes of each curve:

Y = 1 + X^4/X^2-X^4

I got Horizontal = -1 and Vertical = 1, -1
i suppose you mean $\displaystyle y = \frac {1 + x^4}{x^2 - x^4}$ (you should use parentheses).

your horizontal asymptote is fine. as for you vertical asymptotes, what about x = 0?

Quote:

Problem 3
Same as above

Y = 2e^x/e^x-5

I got Horizontal = 0 and Vertical = -1
your vertical asymptote is wrong. remember, you want $\displaystyle e^x - 5 = 0$

for the horizontal, you forgot y = 2. remember you have to take two infinite limits here, $\displaystyle \lim_{x \to \infty}f(x)$ and $\displaystyle \lim_{x \to - \infty}f(x)$

Quote:

Problem 4
Find the limit

Lim x-> positive infinity COSX

I got Lim doesn't exist because COSX is a periodic.
correct, it "oscillates"
• Oct 7th 2007, 09:08 PM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
did you mean $\displaystyle \lim_{T \to \infty} \frac {T^2 + 2}{T^3 + T^2 - 1}$? if so, you are correct

i suppose you mean $\displaystyle y = \frac {1 + x^4}{x^2 - x^4}$ (you should use parentheses).

your horizontal asymptote is fine. as for you vertical asymptotes, what about x = 0?

your vertical asymptote is wrong. remember, you want $\displaystyle e^x - 5 = 0$

for the horizontal, you forgot y = 2. remember you have to take two infinite limits here, $\displaystyle \lim_{x \to \infty}f(x)$ and $\displaystyle \lim_{x \to - \infty}f(x)$

correct, it "oscillates"

Wow thank you a lot. Yeah I think i should have use parenthesis to make myself more clear. You're the best Jhevon:)
• Oct 7th 2007, 09:15 PM
Jhevon
Quote:

Originally Posted by FalconPUNCH!
Wow thank you a lot. Yeah I think i should have use parenthesis to make myself more clear.

you're welcome. and yes, parentheses are important.
Quote:

You're the best Jhevon:)
not really. i let you wait for over an hour for questions i could have answered in less than a minute. sorry about that. i should treat users who attempt their own questions with more respect. good luck, you seem to know your stuff for the most part. just remember, for horizontal asymptotes, you take two infinite limits, one as we go to infinity, one as we go to -infinity. and we have vertical asymptotes for a rational function everywhere the denominator is zero
• Oct 8th 2007, 05:34 AM
FalconPUNCH!
Quote:

Originally Posted by Jhevon
you're welcome. and yes, parentheses are important. not really. i let you wait for over an hour for questions i could have answered in less than a minute. sorry about that. i should treat users who attempt their own questions with more respect. good luck, you seem to know your stuff for the most part. just remember, for horizontal asymptotes, you take two infinite limits, one as we go to infinity, one as we go to -infinity. and we have vertical asymptotes for a rational function everywhere the denominator is zero

Yeah my professor didn't tell us how to find the asymptotes so I tried to the best of my knowledge. Thanks again.