Does the Limit Comparison Test Work for Divergent Integrals?

There are often simpler tricks, like algebraic manipulations or using known limits. In this case, using the known limit of arctan(x) as x goes to infinity would also work. So your summary is: "In summary, the limit comparison test was used to solve the integral \displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}. The comparison function of 1/x was used, and after simplifying, the limit was found to be 1. This showed that the integral diverges, as confirmed by using the known limit of arctan(x) as x goes to infinity."
  • #1
whatlifeforme
219
0

Homework Statement


use limit comparison test.


Homework Equations


[itex]\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}[/itex]


The Attempt at a Solution


I have tried usin 1/x as the comparison function, but when applying the test it
comes out to 0, not an L -> 0 < L < ∞
 
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  • #2
whatlifeforme said:

Homework Statement


use limit comparison test.


Homework Equations


[itex]\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}[/itex]


The Attempt at a Solution


I have tried usin 1/x as the comparison function, but when applying the test it
comes out to 0, not an L -> 0 < L < ∞

Show how you got the limit to be L=0. I get L=1.
 
  • #3
[itex]\displaystyle limit (x->inf) \frac{1/x}{1/\sqrt{x^2 - 1}}[/itex]

[itex]\displaystyle limit (x->inf) \frac{\sqrt{x^2 - 1}}{x}[/itex] = inf/inf

[itex]\displaystyle limit (x->inf) \frac{2x}{\sqrt{x^2 - 1}}[/itex] = inf/inf

[itex]\displaystyle limit (x->inf) \frac{2}{\sqrt[3/2]{x^2 - 1}}[/itex] = 2/inf = 0
 
  • #4
I you try to use l'Hopital on that you are just going to go in circles until you make a mistake and miss a chain rule, like you did. Use algebra to simplify the limit. sqrt(x^2-1)=x*sqrt(1-1/x^2).
 
  • #5
Dick said:
I you try to use l'Hopital on that you are just going to go in circles until you make a mistake and miss a chain rule, like you did. Use algebra to simplify the limit. sqrt(x^2-1)=x*sqrt(1-1/x^2).

i'm sorry I'm lost, and i don't think i left out the chain rule i just didn't include the simplifications in the work above.
 
  • #6
whatlifeforme said:
i'm sorry I'm lost, and i don't think i left out the chain rule i just didn't include the simplifications in the work above.

I don't know what simplifications you made, since you didn't show them, but they aren't right. The first l'Hopital should give you ##\frac{x}{\sqrt{x^2 - 1}}##, the next will give ##\frac{\sqrt{x^2 - 1}}{x}##. Etc, etc.
 
  • #7
how does this look:

1/sqrt(x^2-1) / (1/x)

[itex]\frac{x}{\sqrt{x^2 - 1}}[/itex]

lim (x->inf) [itex]\frac{1}{\sqrt{1-(1/x^2)}}[/itex] = 1

[itex]\displaystyle\int_2^∞ {(1/x) dx}[/itex]

[itex]ln|x| ^{∞}_{2}[/itex]

diverges.
 
  • #8
whatlifeforme said:
how does this look:

1/sqrt(x^2-1) / (1/x)

[itex]\frac{x}{\sqrt{x^2 - 1}}[/itex]

lim (x->inf) [itex]\frac{1}{\sqrt{1-(1/x^2)}}[/itex] = 1

[itex]\displaystyle\int_2^∞ {(1/x) dx}[/itex]

[itex]ln|x| ^{∞}_{2}[/itex]

diverges.

Looks fine. l'Hopital's isn't the best way to handle every limit.
 

What is the Limit Comparison Test?

The Limit Comparison Test is a method of determining the convergence or divergence of a series by comparing it to a known series whose convergence or divergence is already known.

How is the Limit Comparison Test used?

To use the Limit Comparison Test, you must first choose a known series that has the same general behavior as the series in question. Then, you take the limit of the ratio of the terms of the two series. If the limit is a positive, finite number, the two series behave similarly and have the same convergence or divergence. If the limit is 0 or infinity, the two series do not behave similarly and the convergence or divergence of the series in question can be determined.

What is the difference between the Limit Comparison Test and the Ratio Test?

The Limit Comparison Test and the Ratio Test are similar in that they both involve taking the limit of the ratio of the terms of two series. However, the Ratio Test is used to determine the convergence or divergence of a series by comparing it to the geometric series, while the Limit Comparison Test compares a series to a known series with similar behavior.

What is the advantage of using the Limit Comparison Test?

The Limit Comparison Test allows for the determination of the convergence or divergence of a series without having to explicitly find the limit of the series. This can be useful when the series is complex and finding the limit is difficult or impossible.

Can the Limit Comparison Test be used for all series?

No, the Limit Comparison Test can only be used for series with positive terms. It is also only useful when a known series with similar behavior can be identified. In some cases, other convergence or divergence tests may be more appropriate.

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