Question

Check my work please

Answer 1

@mathmale \[Lim (X->\infty) \frac{ 5x^3 - 3x +4 }{ x+11 } => \frac{ \frac{ 5x^3 - 3x +4 }{ x3 } }{ \frac{ x+11 }{ x^3 } } => \frac{ 5 + \frac{ 3 }{ x^2 } + \frac{ 4 }{ x^3 }}{ \frac{ x }{ x^3 } +\frac{ 11 }{ x^3 }} = \frac{ 5+0+0 }{ 0 + 0} \ \frac{ 5 }{ 0 } = 0\]

Answer 2

First I'm going to respond to the question you sent me through the messenger feature of OpenStudy:

My interpretation of the function you typed is as follows:

\[\frac{ x+2 }{ x-3 }\]

Answer 3

Yes that is correct with the lim of X->\[x \rightarrow 3- and x \rightarrow 3+\]

Answer 4

Now suppose you let x get closer and closer to 3 (without touching 3) from BELOW. You might try x=2.9, 2.99, 2.999, 2.9999, and so on. In every case the resulting fraction would be negative.

Now suppose you do the same thing, but approach 3 from above. In every case the resulting fraction would be positive.

If you have a calculator, you might want to try this.

Answer 5

You cannot just substitute x=3, because your fraction would become undefined due to division by zero.

Answer 6

Okay

Answer 7

x -> 3+ does become undef. where as 3- = 1\6. IF i plug them in

Answer 8

Please see

https://www.google.com/webhp?tab=mw&ei=3fcDU_jhLuKSywPe2oDwDg&ved=0CAUQqS4oAg#q=(x%2B2)%2F(x-3)

I've requested a graph of this function. If this site works for you, you'll see that the function is undefined at x=3. Let me know if you can view the graph.

Answer 9

Yes i can, so the graph never touches 3

Answer 10

Yep makes sense

Answer 11

Hint: you can either zoom in or zoom out; it would help you if you were to zoom out. Imagine a vertical line through x=3. That's your vertical asymptote, and yes, you're right: the graph never touches x=3!! See why you can't just plug in x=3?

Answer 12

Now I'm going to look at the last problem y ou've posted. Give me a moment, please.

Answer 13

Okay, but how would I solve the above then? If i cannot plug it in, nor factor it. where do i go?

Answer 14

Just a sec. First:

\[Lim (X->\infty) \frac{ 5x^3 - 3x +4 }{ x+11 } => \frac{ \frac{ 5x^3 - 3x +4 }{ x3 } }{ \frac{ x+11 }{ x^3 } } => \frac{ 5 + \frac{ 3 }{ x^2 } + \frac{ 4 }{ x^3 }}{ \frac{ x }{ x^3 } +\frac{ 11 }{ x^3 }} = \frac{ 5+0+0 }{ 0 + 0} \ \frac{ 5 }{ 0 } = 0\]

Answer 15

Now, to answer your immediate question: Let me look at it once more. Hold, please.

Answer 16

You typed that problem into a message, but did not post it. So I haven't yet read the instructions for this problem. What are they?

Answer 17

Find the lim \[Lim _{X \rightarrow 3-} \frac{ x+2 }{ x-3 } ; Lim _{X \rightarrow 3+} \frac{ x+2 }{ x-3 }\]

Answer 18

(I can only post one problem at a time correct?)

Answer 19

I see. One sided limits. As you approach 3 from the left, the given rational function approaches -infinity; as you appro. 3 from the right, the fn. appr. +infinity. That's it.

Answer 20

Okay that's the answer that was given to us in the example during lecture but wasn't explained. But the graph helped.

Answer 21

While I'd prefer you post only 1 question at a time, it was I who moved a 2nd question on top of a 1st question, because I needed to use Equation Editor. Don't worry about it.

Answer 22

Great.
As much as I'd like to continue, I need a break after having been on my computer most of the day. I look forward to working with you again. All the best to you.

Answer 23

Thank you for your help :) you really did help, and didnt jsut give answers. see you around.