# Limit Comparison Test

1. Mar 13, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
Use direct comparison test or limit comparison test to determine if the integral converges.

2. Relevant equations
$\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$

3. The attempt at a solution

If i were to use the limit comparison test, would these integrals fit the criteria.
** if the positive functions f and g are continues on [a,∞)

Note: what does it mean by positive functions?

limit$_{x->infinity} \frac{f(x)}{g(x)} = L$ 0 < L < ∞

then $\displaystyle\int_a^∞ {f(x) dx}$ and

$\displaystyle\int_a^∞ {g(x) dx}$

both converge or diverge.

f(x) = $\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$

g(x) = $\displaystyle\int_0^6 {\frac{1}{x^2} dx}$

since these functions are not continuous at a=0, then is the limit comparison test not an option here?

If not, how would i go about choosing a function for the direct comparison test?

Last edited: Mar 13, 2013
2. Mar 13, 2013

### Dick

No need to hyperbold on the font. Your integral already has a problem way before infinity. Where is it?

3. Mar 13, 2013

### whatlifeforme

i was stating the theorem from the book. it states must be continuous from [a,∞), where the a=0 in the case, and (sorry forgot to include the comparing function).

$\displaystyle\int_0^6 {\frac{1}{x^2} dx}$ is not continuous at a=0.

f(x) = $\displaystyle\int_0^6 {\frac{dx}{9-x^2}}$

g(x) = $\displaystyle\int_0^6 {\frac{1}{x^2} dx}$

further, i don't think it is stating the integral must be to infinity, just that it must be continuous on, or am i incorrect?

how should i solve with direct comparison test, if easier. please help. test in 2 days.

4. Mar 13, 2013

### Dick

Do direct comparison. The problem with your integral happens at x=3, doesn't it? It's not continuous there. Is 1/(9-x^2) integrable on the interval [0,3]? Try to think of a comparison function on just that interval. Factor 9-x^2.

5. Mar 14, 2013

### whatlifeforme

is (3+x) or (3-x) < (3+x)(3-x) meaning $\frac{1}{3+x} or \frac{1}{3-x} > \frac{1}{9-x^2}$ ??

thus, since

$\displaystyle\int_0^6 {\frac{1}{3+x} dx}$ converges to a finite number.

and $\frac{1}{3+x}> \frac{1}{9-x^2}$

$\displaystyle\int_0^6 {\frac{1}{9-x^2} dx}$ it would conclude that this also converges?

By Direct Comparison Test.

6. Mar 14, 2013

### Dick

The comparison you want is something like $\frac{1}{9-x^2} > C \frac{1}{3-x}$ on the interval 0<x<3 for some constant C. Because $\frac{1}{3-x}$ diverges on that interval allowing you to conclude the original integral diverges. Can you find such a constant C?

7. Mar 14, 2013

### whatlifeforme

i'm sorry. i am still confused. is what i stated above incorrect? the integral is from 0 to 6 not 0 to 3.

what if i choose 9 for c? i just chose that at random.

8. Mar 14, 2013

### Dick

Yes, what you stated before is incorrect. I know your integral is 0 to 6. If you can show it diverges on 0 to 3 then it will diverge on 0 to 6. Why don't you try making a nonrandom choice of C? Use the factorization and figure out what range of values 1/(3+x) takes on [0,3].

9. Mar 14, 2013

### whatlifeforme

are we working with 1/(3-x) or 1/(3+x) or does it matter because we have switched up a few times?

if we are using 1/(3+x), do i need to integrate on [0,3] to find the range of values?

10. Mar 14, 2013

### iRaid

Why don't you just split this integral into two parts and solve for both using limits from the right and left.
$$\int_0^6 \frac{dx}{9-x^{2}}=\lim_{t \to 3^{-}}\int_0^t \frac{dx}{9-x^{2}}+\lim_{t \to 3^{+}}\int_t^6 \frac{dx}{9-x^{2}}$$
See what I'm saying?

11. Mar 14, 2013

### whatlifeforme

if it were, 1/(9+x^2) i would choose 1/x^2 and if it converged i would know the other converged.

the -x^2 is throwing me off in 1/(9-x^2).

(-1/16) < (1/16) when we get into negatives, correct, i know this sounds elementary, but in this application when we are determining smaller and larger functions are negatives always smaller than positives?

12. Mar 14, 2013

### whatlifeforme

i want to know how to solve using a limit comparison test or direct comparison test, not by evaluating.

13. Mar 14, 2013

### Dick

Negatives are always smaller than positives, but when you doing a comparison test you usually want to compare things of the same sign. Just think about it this way. 1/(9-x^2)=(1/(3+x))(1/(3-x)). If you replace the 1/(3+x) with its minimum on [0,3], then you can also replace the '=' sign with '>='.

14. Mar 14, 2013

### iRaid

Oh my bad sorry.

15. Mar 14, 2013

### whatlifeforme

what do you mean by the minimum on [0,3] ? 1/6 would be the minimum, or do you mean integrate on [0,3] ?

16. Mar 14, 2013

### Dick

Yes, I mean 1/6. So 1/(9-x^2)>=(1/6)(1/(3-x)). Does the integral of 1/(3-x) converge on [0,3)?

17. Mar 14, 2013

### whatlifeforme

$\displaystyle\int_0^3 {\frac{1}{3-x} dx}$

$\lim_{b \to 3^-} ln|3-x|]^{0}_{b}$

$\lim_{b \to 3^-} ln|3-b| - ln|3-0| --> ln|3-3| = ln|0| = ∞$

Thus, the smaller diverges, so the original function diverges.

18. Mar 14, 2013

### Dick

Right!

19. Mar 14, 2013

### whatlifeforme

so explain one last time how you got: 1/(9-x^2)>=(1/6)(1/(3-x))

20. Mar 14, 2013

### Dick

It's all in post 13. I factored 1/(9-x^2) and then replaced one of the factors by something smaller.