# Limit as x approaches infinity

1. Sep 5, 2012

### Painguy

1. The problem statement, all variables and given/known data
Use algebraic manipulations to evaluate the limit below.
lim (x^2 +4)/(x+3) X->inf

2. Relevant equations

3. The attempt at a solution
(x^2 +4)/(x+3)
(1+4/x^2)/(1+3/x^2)
1/1=? I My first couple attempts i got wrong so I'm trying again. Am I right this time around?

2. Sep 5, 2012

### Staff: Mentor

You can't do this (above). You divided the numerator by x2 but divided the denominator by x. You can factor the same quantity out of both, or you can do polynomial long division.

Edit: Now I see what you did - you factored x2 out of top and bottom. You have a mistake in the denominator. Instead of factoring x2 out, try factoring x out of num. and denom.

Last edited: Sep 5, 2012
3. Sep 5, 2012

### Painguy

oh that's a typo on my part. looks like i got lost from there
(1+4/x^2)/(1/x+3/x^2)

i would end up with 1/0 as x tends to infinity which is undefined correct?

4. Sep 6, 2012

### Staff: Mentor

No, that's wrong. Like I said already in post #2,

5. Sep 6, 2012

### e^(i Pi)+1=0

Try multiplying both the top and bottom by $\frac{1}{x}$, then evaluate the limit. Or solve it just by looking at it. As x gets bigger and bigger, what happens to the expression?

6. Sep 6, 2012

### Painguy

So it would be infinity? what I'm confused about is why we factor out x and not x^2. Aren't we supposed to factor out the term with the largest degree?

7. Sep 6, 2012

### HallsofIvy

Staff Emeritus
I would say that what you did before, factoring $x^2$ out of both numerator and denominator and then cancelling, to get
$$\frac{1+\frac{4}{x^2}}{\frac{1}{x}+\frac{3}{x^2}}$$
and determinining that the limit does not exist, was correct. Saying the limit "is infinity" is just saying that it does not exist.

It doesn't matter whether you get a finite denominator with the numerator going to infinity or a non-zero numerator with the denominator going to 0.

8. Sep 6, 2012

### mtayab1994

When you have a limit that approaches infinity you always take the highest power in the numerator and divide it by the highest power in the numerator. So you will end up with something like. $$\lim_{x\rightarrow\infty}\frac{x^{2}}{x}$$ It's pretty obvious from here.

9. Sep 6, 2012

### Staff: Mentor

I agree, but I'm not sure that the OP was distinguishing between limits that don't exist because they are infinite, and limits that just plain don't exist, such as this one:
$$\lim_{x \to 0}\frac{1}{x}$$

Having a 1/x term in the denominator falls in the latter category.

Last edited: Sep 6, 2012
10. Sep 6, 2012

### Staff: Mentor

Is that really what you meant to say?
This is simpler, IMO:
$$\lim_{x \to \infty}\frac{x + \text{terms that go to zero}}{1 + \text{terms that go to zero}}$$

11. Sep 6, 2012

### Painguy

So I'm guessing people have different ways of talking about this concept? :P

12. Sep 6, 2012

### mtayab1994

But in this case you are aloud to say that. You can cancel out with the x's and you be left with the limit of x as x approaches infinity and that's really simple.

13. Sep 7, 2012

### Staff: Mentor

What you said:
If that's what you really meant to say, it makes no sense, thus my comment.

You also said:
You don't actually "end up" with this, as it is the indeterminate form [∞/∞]. As you said in a later post, you need to do some more work.

14. Sep 7, 2012

### Staff: Mentor

I'm not sure what you mean. If you mean the different ways that a limit can fail to exist, it can be because the expression is unbounded (has ∞ or -∞ as its limit) or no limit exists at all, such as $\lim_{n \to \infty} (-1)^n$.

If you mean different approaches to this problem, different approaches should produce the same value for the limit. For a rational function like the one you posted, the most straightforward way is to find the smaller leading exponent of the numerator and denominator, and divide numerator and denominator by the variable raised to that power.

15. Sep 7, 2012

### mtayab1994

But x^2/x becomes x/1 and the limit of that is infinity.

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