Question

I need some quick help on limits questions- I'm about to take an exam and I want to be understand these questions that I got wrong. It's only three questions, and I'm going to attach them as images below. The selected answers are what I originally chose, which is wrong. Thanks a ton ahead of time!

Answer 1

Here's the first one

Answer 2

Here's the second one

Answer 3

And here is the last one. I'll give a medal to whoever explains these to me please <3

Answer 4

Or more than 1 if that's possible?

Answer 5

Ugh, multiple choice... Hints only.

First: Your choice isn't wrong, but it's missing the other choice that follows directly from the one you picked.

Second: I assume you factorized the denominator and canceled the factors of \(x+3\), which isn't necessarily wrong, but what happens in the original function if \(x=-3\)?

Third: Long division yields
\[\frac{7x^3-3x^2+3x}{-8x^2+4x+3}=-\frac{7}{8}x-\frac{1}{16}+\text{some remainder}\]

Answer 6

Too start, thank you for the input! :)
Secondly, I just want to check if what I'm getting from your help is right (which I really appreciate btw). I think I have the second two down, but I'm still a bit confused with the first one.
For the first question, I think I understand you're telling me it's d, or all statements are true, but I still don't get why.
For the second question, the equation would be discontinuous at both x=1 and x=-3, and this would be the case because even though -3 is a discontinuity, it's still removable?
And lastly for the third question, would that mean the answer is negative infinity?
@SithAndGiggles

Answer 7

Btw, I'll give you another medal if I can when you help me out more. I really appreciate it!

Answer 8

First: Yes, all of these are true. (I) and (III) mean the same thing, given that \(f(x)\) is odd and \(f(x)\to-5\) as \(x\to-\infty\). You also know that \(f(x)\) is continuous, so it has no vertical asymptotes.

Second: Yes, a removable discontinuity is still a discontinuity.

Third: Yes. The remainder term disappears for large values of \(x\), so you're left with a polynomial.
\[\lim_{x\to\infty}\frac{7x^3-3x^2+3x}{-8x^2+4x+3}=\lim_{x\to\infty}\left(-\frac{7}{8}x-\frac{1}{16}\right)=-\infty\]