# Limit comparison test

1. Mar 17, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
use limit comparison test.

2. Relevant equations
$\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}$

3. 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 < ∞

2. Mar 17, 2013

### Dick

Show how you got the limit to be L=0. I get L=1.

3. Mar 17, 2013

### whatlifeforme

$\displaystyle limit (x->inf) \frac{1/x}{1/\sqrt{x^2 - 1}}$

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

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

$\displaystyle limit (x->inf) \frac{2}{\sqrt[3/2]{x^2 - 1}}$ = 2/inf = 0

4. Mar 17, 2013

### Dick

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. Mar 17, 2013

### whatlifeforme

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. Mar 17, 2013

### Dick

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. Mar 17, 2013

### whatlifeforme

how does this look:

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

$\frac{x}{\sqrt{x^2 - 1}}$

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

$\displaystyle\int_2^∞ {(1/x) dx}$

$ln|x| ^{∞}_{2}$

diverges.

8. Mar 17, 2013

### Dick

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

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