Question

The limit as x approaches infinity of the integral from x to 2x of dt/t

Answer 1

What? I think I know what you're talking about, and I htink you're going to use L'hopitals rule, but can you re-write it using LaTex?

Answer 2

\[\lim_{x \rightarrow \infty}\int\limits_{x}^{2x}\frac{ dt }{ t }\]

Answer 3

Do you know the answer?

Answer 4

Because I believe the integral simply diverges.

Answer 5

Like the limit doesn't exist.

Answer 6

the answer is just ln2, but I don't know how my teacher got that

Answer 7

Hmm... One moment.

Answer 8

Okay

Answer 9

I am seriously stuck. I would love to know how yo approach this.

Answer 10

I multiplied the top and bottom by x and applied L'hospitals rule but that doesn't help.

Answer 11

I'm stuck too, I've been stuck on this problem for a while now. Thanks anyways :)

Answer 12

Well zzr0ck3r is replying. He is pretty good.

Answer 13

nah just typing so i can see:)

Answer 14

im stuck

Answer 15

X) .

Answer 16

I really hope he can help me out , I have a test tomorrow and I need to figure this out

Answer 17

Well I tried to multiply the top and bottom by x in order to apply l'hospitals rule.

Answer 18

But the problem is that we still get infinity

Answer 19

So I can't really do anything. Applying L'hospital's rule again doesn't help either.

Answer 20

But can't l'hopital's rule be applied more than once ?

Answer 21

Yeah I know. But that makes the situation worse.

Answer 22

Yeah that's true...

Answer 23

ok got it

Answer 24

Does it involve splitting up the integral? Then I think I got it too.

Answer 25

\[\int_{x}^{2x}\frac{1}{t}dt=-\int_{0}^{x}\frac{1}{t}dt+\int_{0}^{2x}\frac{1}{t}dt=\\-\ln(x)+\ln(0)+\ln(2x)-\ln(0)=-\ln(x)+\ln(2)+\ln(x) = \ln(2)\]

Answer 26

im not sure if you can do that but it looks good to me:)

Answer 27

ln(0) is undefined.

Answer 28

ahh we could have taken the integral to 1

Answer 29

Pick some other value like 2 :P .

Answer 30

yeah...

Answer 31

ok have fun, tv show with the wife:)

Answer 32

What about the limit though?

Answer 33

But okay have fun X) .

Answer 34

NVM I got the limit part too.

Answer 35

@jossy04 : did you get what he did?

Answer 36

So the way zzr0ck3r did it is wrong ??

Answer 37

No it's right haha,

Answer 38

Yeah I get what he did

Answer 39

But when he split up the intergal he used the wrong limits.

Answer 40

Okay haha thanks :)

Answer 41

Because ln(0) is undefined at 0.

Answer 42

He should have split it up at like 1 or 2 or something non negative or 0.

Answer 43

Ohh okay then .

Answer 44

@Dido525 do you happen to know how to solve this one too ? \[\lim_{x \rightarrow 1}\frac{ \int\limits_{1}^{x}\cos tdt }{ x^{2}-1 }\]

Answer 45

Yep I can do that.

Answer 46

Can you help me please ? :)

Answer 47

First notice that if that thing is actually approaching 1 note how the function become 0/0 right?

Answer 48

Because the integral doesn't move anywhere it's just going to be a 0.

Answer 49

Does that make sense?

Answer 50

Yeah it makes sense .

Answer 51

Alright we know l'hospitals rule can be applied if the limit approaches 0/0 or infinity/infinity

Answer 52

Do you know the fundamental theorem of Calculus?

Answer 53

People! What are you doing? Just evaluate it directly. No l'Hopital, no scary limits, no splitting up.

For some x > e (really, just pick a positive number. I used 'e' just for fun. Convergence doesn't care about some small finite area at the beginning.)

\(\int\limits_{x}^{2x}\dfrac{1}{t}\;dt = ln(t)|_{x}^{2x} = ln(2x) - ln(x) = ln\left(\dfrac{2x}{x}\right) = ln(2)\)

The result is independent of x and the limit is of no consequence.

Answer 54

I can't believe I didn't hink of Logarithmic properties. I was stupid :( .

Answer 55

So L'hospitals rule says if we get a limit in the form 0/0 we take the limits of the derivatives in the numerator and denominator.

Answer 56

Do you understand that so far?

Answer 57

What tkhunny did was so much easier haha
Yeah I'm understanding .
Thanks @tkhunny :)

Answer 58

Allright. Do you get how the fundamental theorem of Calculus works?

Answer 59

Yeah I get how it works .

Answer 60

Alright, so we have that x in that integral. We substitute x instead of a t inside that cosine and multiply by the derivative of x which is just 1.

Answer 61

The derivative of the bottom is just 2x

Answer 62

So we get our limit as: