Question

Find the limit of the function algebraically. see attached.

Answer 1

@zepdrix

Answer 2

how do you know that? I'm very new at this. This is the first time I've worked with limits.

Answer 3

The denominator will go to +0 from either direction, while the numerator goes to -6. That gives a small negative number over an infinitely small positive number, making the magnitude |infinite|, so the limit is:

negative infinity.

Answer 4

That's not an answer choice. These are the answer choices for what the limit is 6

0

-6

Does not exist

Answer 5

It is negative infinity. The graph looks like this:

Answer 6

Hmm yah it's approaching \(\large -\infty\).
Strange that they didn't give you that answer choice. D:

Answer 7

Should I choose does not exist?

Answer 8

Yes, that's probably your best bet :)

Answer 9

Well, that's like saying, "What's my best wrong choice?"

Answer 10

okay I will. I have another problem very similiar to this one.

Answer 11

\[\large \lim_{x \rightarrow 0}\frac{1}{x}\]

emerson, do you understand why this limit approaches infinity?
If you understand this one, it might give you an idea of how these work.

Answer 12

I just feel as if since there is an x on the bottom and the x will be zero, that it wouldn't exist. Not really.. I know how limits work, but it's that there is a 0 on the bottom that confuses me.

Answer 13

Ok let's take a number really close to zero, from the right side.
So let's let \(\large x=\dfrac{1}{99999}\)

That's pretty close to zero right?

Answer 14

If we plug this number into our function, it might give us an idea what's happening since this is "close" to zero.

\[\large \frac{1}{(x)}\qquad\rightarrow\qquad \frac{1}{\left(\dfrac{1}{99999}\right)} \qquad=\qquad 99999\]

It's producing a really really big number when we plug in a value close to zero.

Answer 15

So then when you put in a positive number that is really close to zero it will be approaching positive infinity and when you put in negative number very close to zero it will approach negative infinity?

Answer 16

Yes! Good observation! :)

Answer 17

Ok now I see the mistake we made in the last problem,
Let's break up the fraction and maybe you'll see what's going on.
\[\large \lim_{x \rightarrow 0}\frac{9+x}{x^3} \qquad=\qquad \lim_{x \rightarrow 0}\frac{9}{x^3}+\frac{x}{x^3}\qquad=\qquad \lim_{x \rightarrow 0}\frac{9}{x^3}+\frac{1}{x^2}\]

Answer 18

These steps make sense so far?

Answer 19

Yes. You separated the one fraction into two and the second one, you were able to cancel out an x.

Answer 20

Using rules of limits, we can rewrite this as the sum of limits.\[\large \lim_{x \rightarrow 0}\;\frac{9}{x^3}+\lim_{x \rightarrow 0}\;\frac{1}{x^2}\]

Answer 21

The limit on the right is not a problem.
Whether we approach zero from the left or the right, our function is approaching the same value either way, \(\large \infty\).

That's due to the square which eliminates the issue with approaching from the negative side.

Answer 22

But what's happening with the first limit?
If we approach from the left, what is our function approaching?
If x approaches zero from the right, what is our function approaching?

Answer 23

`The limit on the right is not a problem.`
I meant the 2nd limit that I wrote out. I should have made that clearer by not using the word `right`. lol

Answer 24

Uh oh, silence!
Confused by some of that? c:

Answer 25

yeah..ok

Answer 26

If it approaches from the left it will approach negative infinity and from the right positive infinity?

Answer 27

Yah, so if it says \(\large x\rightarrow 0\) without specificying from the left or right, then we can determine that the limit does not exist.
~Because the left and right limit do not agree.

Answer 28

Okay, that makes sense. Thank you!