# Math Help - How do I find this limit

1. ## How do I find this limit

$x\neq {-3,1}$

a) $\lim_{x \to -3+}g(x) = \frac{5x^2}{x^2+2x-3}$

b) $\lim_{x \to 1-}g(x) = \frac{5x^2}{x^2+2x-3}$

I can solve these by subbing in values just a bit greater than 3 and using the same technique for b), but how do I solve these algebraically. There were two questions before these where $x~\rightarrow$ $\infty$ (same function) and I just divided the top and bottom by $x^2$ to get easily solvable limits. I tried this for these two and I get a denominator of 0 - so it doesn't work.

2. ## Re: How do I find this limit

Why not just look at it? Maybe factor the denominator? What does "Vertical Asymptote" mean? According to the defininition of a limit, can you demonstrate that there is a value that never will be exceeded if you stay close enough? Can you demonstrate that any value selected can be exceeded?

Note: Please don't consider l'Hospital. It's NOT an Indeterminate Form! (around -3 or 1, anyway.)

3. ## Re: How do I find this limit

Originally Posted by terrorsquid
$x\neq {-3,1}$

a) $\lim_{x \to -3+}g(x) = \frac{5x^2}{x^2+2x-3}$

b) $\lim_{x \to 1-}g(x) = \frac{5x^2}{x^2+2x-3}$

I can solve these by subbing in values just a bit greater than 3 and using the same technique for b), but how do I solve these algebraically. There were two questions before these where $x~\rightarrow$ $\infty$ (same function) and I just divided the top and bottom by $x^2$ to get easily solvable limits. I tried this for these two and I get a denominator of 0 - so it doesn't work.
Are you allowed to use L'Hospital's Rule? If so, use it. If not, try simplifying using Partial Fractions...

4. ## Re: How do I find this limit

Originally Posted by TKHunny
Why not just look at it? Maybe factor the denominator? What does "Vertical Asymptote" mean?
(around -3 or 1, anyway.)
I tried factoring, $\frac{5x^2}{(x+3)(x-1)}$ ? 0 denominator.

There are vertical asymptotes at x = 1 and x = -3

I can see that the function goes to $-\infty$ but only by subbing in values a bit greater than -3* like -2.9999 etc.

Originally Posted by TKHunny
Can you demonstrate that any value selected can be exceeded?
$\forall x \in \mathbb{R},~ \exists y \in \mathbb{R} \{x|y>x\}$ ?

Originally Posted by TKHunny
According to the defininition of a limit, can you demonstrate that there is a value that never will be exceeded if you stay close enough?
?

5. ## Re: How do I find this limit

Originally Posted by Prove It
Are you allowed to use L'Hospital's Rule? If so, use it. If not, try simplifying using Partial Fractions...
This is an old practice sheet I am going over; we hadn't covered L'Hospital's rule at that stage. Partial fractions... ok let me look that up (I tried dividing top and bottom by x^2 like I did with the $x\rightarrow \infty$ versions of this problem but ran into 0 denominator trouble again.)

6. ## Re: How do I find this limit

Lovely notation, but you seem to have forgotten the problem at hand.

1) You missed an important question. What does "Vertical Asymptote" mean?

2) Since we are approaching -3 from the right, pick a value that you think might be a limit as x approaches -3. Say, -10. Can you find a value of x, near -3 on the positive side, for which f(x) < -10? Okay, after demonstrating for -10, try -1000. Okay, how about -100,000? At some point, we should conclude that since our selection is arbitrary, we are done. There is NO such value that we cannot exceed. What implication have we then for the limit?

3) Rather than try to prove that there is NO limit, one may wish to prove that the function is unbounded. It is very similar, but a little different flavor. Essentially, it is this: Can we find a value, say "L", that once we pass it, we'll never come back by getting closer to -3?

4) You don't seem to have listened to me about l'Hospital, either. It has no application around -3 or 1.

7. ## Re: How do I find this limit

Originally Posted by TKHunny
1) You missed an important question. What does "Vertical Asymptote" mean?
It means y is not defined at these values for x.

Originally Posted by TKHunny
Since we are approaching -3 from the right, pick a value that you think might be a limit as x approaches -3. Say, -10. Can you find a value of x, near -3 on the positive side, for which f(x) < -10? Okay, after demonstrating for -10, try -1000. Okay, how about -100,000? At some point, we should conclude that since our selection is arbitrary, we are done. There is NO such value that we cannot exceed. What implication have we then for the limit?
Isn't this just what I was doing by subbing in values into my calculator? Or do you mean setting $\frac{5x^2}{x^2+2x-3} = -10$ etc. and solving for x. That seems like a very long method

Originally Posted by TKHunny
Rather than try to prove that there is NO limit, one may wish to prove that the function is unbounded. It is very similar, but a little different flavor. Essentially, it is this: Can we find a value, say "L", that once we pass it, we'll never come back by getting closer to -3?
I suspect we can, but the only way I seem to know how to see this is by trial and error with the calculator , i.e., my problem.

8. ## Re: How do I find this limit

Originally Posted by terrorsquid
It means y is not defined at these values for x.

Isn't this just what I was doing by subbing in values into my calculator? Or do you mean setting $\frac{5x^2}{x^2+2x-3} = -10$ etc. and solving for x. That seems like a very long method

I suspect we can, but the only way I seem to know how to see this is by trial and error with the calculator , i.e., my problem.
Or you could use Partial Fractions, like you were instructed...

9. ## Re: How do I find this limit

Originally Posted by terrorsquid
It means y is not defined at these values for x.
No, "vertical asymptote" does NOT mean that. For example, the function [tex]f(x)= \frac{x^2- 4}{x- 2}[/itex] is not defined at x= 2 but does not have a vertical asymptote there.

Isn't this just what I was doing by subbing in values into my calculator? Or do you mean setting $\frac{5x^2}{x^2+2x-3} = -10$ etc. and solving for x. That seems like a very long method
Yes, the second is what TKHunny is talking about. But it is not a method for finding the limit, it is a method for "checking" what you think might be a limit.

I suspect we can, but the only way I seem to know how to see this is by trial and error with the calculator , i.e., my problem.
Typically, one of the very first things that people learn about limits of fractions is:
"If $\lim_{x\to a} g(x)= 0$ and $\lim_{x\to a} f(x)$ is NOT 0, then $\lim_{x\to a}\frac{f(x)}{g(x)}$ does not exist".

10. ## Re: How do I find this limit

It's not a long method at all. It's called EXPLORATION.

11. ## Re: How do I find this limit

Originally Posted by Prove It
Or you could use Partial Fractions, like you were instructed...
I still have x+3 as the denominator of one of the fractions though, and therefore can't sub in the value -3 and am in the same predicament. I couldn't seem to figure out how partial fractions help in this situation.

12. ## Re: How do I find this limit

find g'(x) in order to find the maximum and min point, also you need to find the vertical asymptote. Make a sketch of g(x), then you will know the answer.
I think this is a way to do it if you have learnt derivative, but ignore this if you haven't.

13. ## Re: How do I find this limit

I can solve these by subbing in values just a bit greater than 3 and using the same technique for b), but how do I solve these algebraically. There were two questions before these where (same function) and I just divided the top and bottom by to get easily solvable limits. I tried this for these two and I get a denominator of 0 - so it doesn't work.
I just want to be clear... Your know how to solve these limits, but you want to know how to solve them algebraically?

If so, it is not possible. When you solve a limit algebraically, you're solving as it approaches the number from both sides. In this case, both are vertical asymptotes, and both are approaching negative infinite and positive infinite from opposite directions, meaning the limit does not exist. It is just like the other limit question you posted.

In short, you won't be able to solve it algebraically. However, the actual question asks for the limit as we get closer from only one side, so there is a limit (which you said you know how to do); you just cannot solve it algebraically.

I hope this helps!

-Nathan

14. ## Re: How do I find this limit

There were two questions involving this function that I did before the two I posted: $\lim_{x\to+\infty}$ and $\lim_{x\to-\infty}$
and I was able to solve them algebraically as follows:

$\frac{5x^2}{x^2+2x-3}$

divide top and bottom by $x^2$

$= \frac{5}{1+\frac{2}{x}-\frac{3}{x^2}}$

$\lim_{x\to+\infty}\frac{5}{1+\frac{2}{x}-\frac{3}{x^2}} = 5$

I assumed there was a similar method for the two questions I posted in my first post that I couldn't see is why I asked. I solved them manually by subbing in values with my calculator.

15. ## Re: How do I find this limit

The difference between the original two questions and those two questions is that when you're finding the limit as x approaches negative or positive infinite, you may only be approaching from one side, but that's because you can't approach it from any other side. There's no such thing as infinite times 2. In the original two questions, you could approach from both sides, but they only specified from one side.

As to these two new questions, there's an even simpler way of solving them. When you're taking the limit as x approaches positive/negative infinite, you're actually finding the horizontal asymptotes. To find these, you simply look at the coefficients of the term with the highest degree (in this case, x^2). The coefficients are 5 and 1, so 5/1 is 5.

To make this more clear, if you squared a very large number, and then added that same number, the addition of that same number would be pointless; it would be like adding 1 to 1000000. It just doesn't affect it enough to really care about it. That's why you only look at the term with the highest degree.

Hopefully this all made sense and helps!

-Nathan

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