Evaluate lim as x→3: (x/x-3) Int( sint/t dt )

[Jbjohnson15]

Homework Statement

Evaluate the lim as x approaches 3 of (x/x-3) times the integral from 3 to x of (sint/t)dt

Homework Equations

The Attempt at a Solution

 

[micromass]

So, what did you already do to solve the problem?
How do you usually solve "0/0" situations?

 

[Jbjohnson15]

I tried to evaluate this using the substitution method: making u=sint and du=costdt. Then that created a big jumbled incorrect mess. I don't know how to attack this problem.

 

[micromass]

One of the problems with this particular integral, is that you cannot solve it. So don't bother with solving this integral, it won't work. But there is a way to make this integral disappear: differentiate it. Now, I wonder, is there a way that you can solve a limit using derivatives...

 

[Jbjohnson15]

[f(x+h)-f(x)]/h ? I'm really not sure. What would be the h? 3? I'm so lost.

 

[micromass]

How would you resolve a "0/0"-situation with the aid of derivatives?

Spoiler

L'hopital rule

 

[Jbjohnson15]

L'hospital rule: so take the derivative of the top and the bottom of (x/x-3), thus giving you 1/1 times the integral. How will this make the integral disappear?

 

[micromass]

Your limit is

$$\lim_{x\rightarrow 3}{\frac{x\int_3^x{\frac{\sin(t)}{t}dt}}{x-3}}$$

Try applying l'Hopitals rule on that. You are correct that the integral will not disappear, but the limit will become simpler...

 

[Jbjohnson15]

Ok, so after applying L'Hopitals rule, I ended up with the limit as x approaches 3 of [sin(x)/x + sinx]. That can't be right. I'm sorry that I am mathematically incompetent.

 

[micromass]

no, that can't be right, there should still be an integral in there...
For ease, define

$$F(x)=\int_3^x{\frac{\sin(t)}{t}dt}$$

What is $$F^\prime(x)$$ (this is basically the fundamental theorem of calculus).

Now you want to calculate the derivative of xF(x). How would you do this? (hint: product rule)

 

[Jbjohnson15]

According to the fundamental theorem of calculus, F'(x) equals f(x) or sin(x)/x. Now, upon calculating the derivative of xF(x), I use the product rule which will be x'F(x) + xF'(x). That gives me the integral from 3 to x of sin(t)/t dt + sinx.

 

[micromass]

Yes, that is correct. Now take the limit of that expression. What do you get?

 

[Jbjohnson15]

Well, x is approaching 3, if I plug in 3, the integral will be from 3 to 3 making it zero. Then I'm left with sin(3)...

 

[micromass]

Yes, sin(3) seems to be the correct answer.

 

[Jbjohnson15]

I cannot thank you enough for your patience and kindness. I truly appreciate your help. Thank you and God bless!