# How should i prove this?

• Nov 18th 2009, 06:55 PM
xcluded
How should i prove this?
I'm stuck after i tried to integrate by parts.

$\displaystyle \lim x \int_x^1 \frac{cos t}{t^2} dt = 1$

The limit is actually as $\displaystyle x \rightarrow 0^+$ , I can't seemed to key in the input.

(Nod)
• Nov 18th 2009, 07:22 PM
RockHard
Quote:

Originally Posted by xcluded
I'm stuck after i tried to integrate by parts.

$\displaystyle \lim x \int_x^1 \frac{cos t}{t^2} dt = 1$

The limit is actually as $\displaystyle x \rightarrow 0^+$ , I can't seemed to key in the input.

(Nod)

I wonder if the comparison test would work, which you find a function similar to
the function which in this case you can do, someone correct me if I am wrong
since cosine is bounded by -1 and 1
$\displaystyle -1\le cos(t)\le1$
so then
$\displaystyle \frac{-1}{t^2}\le \frac{cos(t)}{t^2}\le \frac{1}{t^2}$
so in comparsion, you can say you function behaves like the function
$\displaystyle \frac{1}{t^2}$, if you dont know the comparsion test, look it up it can prove helpful, if you find a function larger than your original then you the function you were given which is small will also converge, if you find a function smaller than the original diverges then the original will diverge, this does not apply to vice versa so now you can do

$\displaystyle \lim_{x\to0+}\int_x^1\frac{1}{t^2}$

$\displaystyle \lim_{x\to0+}-\frac{1}{t}\|_x^1$

you should be able to take it from here
• Nov 18th 2009, 07:34 PM
Drexel28
Quote:

Originally Posted by xcluded
I'm stuck after i tried to integrate by parts.

$\displaystyle \lim x \int_x^1 \frac{cos t}{t^2} dt = 1$

The limit is actually as $\displaystyle x \rightarrow 0^+$ , I can't seemed to key in the input.

(Nod)

Is this $\displaystyle \lim_{x\to0^+}x\int_x^1\frac{\cos(z)}{z^2}dz$ or $\displaystyle \lim_{x\to0^+}\int_x^1\frac{\cos(z)}{z^2}dz$

Seeing your asnwer I assume it's the first.

I hate using it, but I assume it's how this is to be done

$\displaystyle \lim_{x\to0^+}\frac{\int_x^1\frac{\cos(z)}{z^2}dz} {\frac{1}{x}}$

Both the denominator and numerator approach infinity as $\displaystyle x\to0^+$. Apply L'hopital's rule and...done.
• Nov 18th 2009, 08:07 PM
xcluded
Quote:

Originally Posted by RockHard
I wonder if the comparison test would work, which you find a function similar to
the function which in this case you can do, someone correct me if I am wrong
since cosine is bounded by -1 and 1
$\displaystyle -1\le cos(t)\le1$
so then
$\displaystyle \frac{-1}{t^2}\le \frac{cos(t)}{t^2}\le \frac{1}{t^2}$
so in comparsion, you can say you function behaves like the function
$\displaystyle \frac{1}{t^2}$, if you dont know the comparsion test, look it up it can prove helpful, if you find a function larger than your original then you the function you were given which is small will also converge, if you find a function smaller than the original diverges then the original will diverge, this does not apply to vice versa so now you can do

$\displaystyle \lim_{x\to0+}\int_x^1\frac{1}{t^2}$

$\displaystyle \lim_{x\to0+}-\frac{1}{t}\|_x^1$

you should be able to take it from here

Hmm what about the x outside the integral ?
• Nov 18th 2009, 08:09 PM
xcluded
Quote:

Originally Posted by Drexel28
Is this $\displaystyle \lim_{x\to0^+}x\int_x^1\frac{\cos(z)}{z^2}dz$ or $\displaystyle \lim_{x\to0^+}\int_x^1\frac{\cos(z)}{z^2}dz$

Seeing your asnwer I assume it's the first.

I hate using it, but I assume it's how this is to be done

$\displaystyle \lim_{x\to0^+}\frac{\int_x^1\frac{\cos(z)}{z^2}dz} {\frac{1}{x}}$

Both the denominator and numerator approach infinity as $\displaystyle x\to0^+$. Apply L'hopital's rule and...done.

i got this

$\displaystyle \lim_{x\to0^+}\frac{x^2 sinx - 2x cos x}{x^2}$

but never got it equal to 1.

i did something wrong?
• Nov 18th 2009, 08:32 PM
redsoxfan325
Quote:

Originally Posted by xcluded
i got this

$\displaystyle \lim_{x\to0^+}\frac{x^2 sinx - 2x cos x}{x^2}$

but never got it equal to 1.

i did something wrong?

$\displaystyle \frac{d}{dx}\left[\int_x^1\frac{\cos t}{t^2}\,dt\right]=-\frac{\cos x}{x^2}$

$\displaystyle \frac{d}{dx}\left[\frac{1}{x}\right]=-\frac{1}{x^2}$

So...
• Nov 18th 2009, 08:38 PM
xcluded
Quote:

Originally Posted by redsoxfan325
$\displaystyle \frac{d}{dx}\left[\int_x^1\frac{\cos t}{t^2}\,dt\right]=-\frac{\cos x}{x^2}$

$\displaystyle \frac{d}{dx}\left[\frac{1}{x}\right]=-\frac{1}{x^2}$

So...

OH ! Ok i did something terribly wrong earlier.
Thanks !
• Nov 18th 2009, 09:00 PM
RockHard
Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP
• Nov 18th 2009, 09:04 PM
xcluded
Quote:

Originally Posted by RockHard
Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP

Yea my lecturer used the comparison test to do it similar question too.
By the way , do i need to compare once only ? As in ,

$\displaystyle \lim_{x\to0+}\int_x^1\frac{1}{t^2} = 1$

Hence the whole thing converges to 1 ?
• Nov 18th 2009, 09:07 PM
redsoxfan325
Quote:

Originally Posted by RockHard
Was my idea somewhat? I am hoping to make some sort of progress here, lol. So If I ever need correct, feel free too even if your the OP

You would need to squeeze it on both sides, not just one.

$\displaystyle \lim_{x\to0^+}x\int_x^1-\frac{1}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{ \cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\int_x^1\frac{1}{t^2 }\,dt$

So $\displaystyle \lim_{x\to0^+}x\cdot\frac{x-1}{x}\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq\lim_{x\to0^+}x\cdot\frac{1-x}{x}$

So all this tells you is that $\displaystyle -1\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1$

Or maybe since it goes to zero from the positive side, $\displaystyle 0\leq\lim_{x\to0^+}x\int_x^1\frac{\cos t}{t^2}\,dt\leq1$