# Thread: Need Calculus assistance with several problems! I have a date tonight!

1. ## Need Calculus assistance with several problems! I have a date tonight!

I am very sorry with being so urgent, however I am having terrible trouble with my Calculus homework - can't figure it out at all and I am working for hours at it.

I have several problems that I may need help explaining. I would prefer not to just get the answer, and so would prefer just hints for now, however I really want/need to get full credit on this assignment, and I have a date set for 7 PM tonight.

The first problem I have issues with is:

Find the limit of x approaches negative infinity of the function: ((10-15x)/(9+x)) + ((9x^2 + 13)/(12x-5)^2)

2. You could rewrite as:

$\displaystyle \frac{-711x^{3}+2121x^{2}-1312x+367}{144x^{3}+1176x^{2}-1055x+225}$

Now, keep in mind that when finding the limit of a rational expression as $\displaystyle x\rightarrow{\pm\infty}$, you can ignore everything except the highest powers.

3. Hello,

Originally Posted by galactus
You could rewrite as:

$\displaystyle \frac{-711x^{3}+2121x^{2}-1312x+367}{144x^{3}+1176x^{2}-1055x+225}$

Now, keep in mind that when finding the limit of a rational expression as $\displaystyle x\rightarrow{\pm\infty}$, you can ignore everything except the highest powers.
Wouldn't it be better to write this way :

$\displaystyle \frac{10-15x}{9+x}+\frac{9x^2 + 13}{(12x-5)^2}=\frac{10-15x}{9+x}+\frac{9x^2 + 13}{144x^2-120x+25}$

?

4. That would seem to give me -4.93, which doesn't work when I try to input the answer.

5. Originally Posted by Pikeman85
That would seem to give me -4.93, which doesn't work when I try to input the answer.
Because it's approximative ?
Where did you put your answer ?

6. I put it into the math homework submitter thing, WebCT.

This time I tried -4.9375, and still got incorrect.

7. Originally Posted by Pikeman85
I put it into the math homework submitter thing, WebCT.

This time I tried -4.9375, and still got incorrect.
Then try -15+9/144=-239/16

It's better keeping the exact values

8. Thank you! How did you get that?! What was I doing wrong? I thought I was using the values correctly. EDIT: I realized that I used a positive 15 and you a negative, would that be because it is approaching negative infinity? If so, that was such a stupid error on my part as I had gotten 15.06... earlier.

And now time for the next question (there's a lot, so feel free to help only as much as you want to guys. I might start coming here frequently just so I can understand this stuff better)

((6x + 10)/(x - 7)) * ((7x - 2)/(-x - 8))

I got this one, was relatively easy after figuring exactly how I went wrong on the last one

9. I wrote it that way with a common denominator so we could see the coefficients of the highest powers are

-711/144 = -79/16 = -4.9375

10. ((6x + 10)/(x - 7)) * ((7x - 2)/(-x - 8))
Is this another limit as x--> -infinity?. If so, did you get -42?.

11. Now this next one has been giving me trouble.

It is as the limit approaches positive infinity, of sqrt [x^2 + 8x - 10] - x

I tried to figure this one out by multiplying it by sqrt [x^2 + 8x - 10] + x, but this did not seem to work out.

12. Yes, I got -42 for that one. It was quite easy after the first one in that sequence, I was just daunted by it.

13. Originally Posted by Pikeman85
Now this next one has been giving me trouble.

It is as the limit approaches positive infinity, of sqrt [x^2 + 8x - 10] - x

I tried to figure this one out by multiplying it by sqrt [x^2 + 8x - 10] + x, but this did not seem to work out.
\displaystyle \begin{aligned} \sqrt{x^2+8x-10}-x &=(\sqrt{x^2+8x-10}-x) \cdot \frac{\sqrt{x^2+8x-10}+x}{\sqrt{x^2+8x-10}+x}\\ &=\frac{(x^2+8x-10)-x^2}{\sqrt{x^2+8x-10}+x} \\ &=\frac{8x-10}{\sqrt{x^2+8x-10}+x} \\ &=\frac{8-\frac{10}{x}}{\sqrt{1+\frac 8x-\frac{10}{x^2}}+1} \end{aligned}

Above, it tends to 8, below, it tends to 2..

14. I believe the help given here might be illegal. Let the moderators investigate. I know what the WebCT system is and how it works...

15. Neither 8 nor 2 worked, and unbounded isn't an option.

As I said, I tried to calculate it out with the equation I specified.

Page 1 of 3 123 Last