L'Hospital's Rule and Integration

[_N3WTON_]

Homework Statement

lim (x→0) of the definite integral (from 1 to 1+x) of (cos(t))/t all over x

Homework Equations

Integration by parts
L'hospital's rule

The Attempt at a Solution

I believe that I must first solve the definite integral, and then take this result and use L'hospital's rule in order to solve the equation. However, using integration by parts on the integral has lead me nowhere. I was hoping somebody could point me in the correct direction. Thanks.

 

[Ray Vickson]

Homework Statement

lim (x→0) of the definite integral (from 1 to 1+x) of (cos(t))/t all over x

Homework Equations

Integration by parts
L'hospital's rule

The Attempt at a Solution

I believe that I must first solve the definite integral, and then take this result and use L'hospital's rule in order to solve the equation. However, using integration by parts on the integral has lead me nowhere. I was hoping somebody could point me in the correct direction. Thanks.

You are mis-using l'Hospital's rule. Go back and review it; you will soon see what is happening.

 

[_N3WTON_]

I know how to use the rule and I just assumed that it would be needed here, the real problem I am having is with the definite integral. Can I just write it in terms of 'x' using The Second Fundamental Theorem?

 

[slider142]

Never mind. The reply below has the better approach.

 

[gopher_p]

Given $F(x)=\int_1^{1+x}\frac{\cos t}{t}\ dt$, you have correctly identified that $\lim_{x\rightarrow 0}F(x)=0$ and that l'Hopital's rule might prove useful in evaluating $\lim_{x\rightarrow 0}\frac{F(x)}{x}$.

Might it be easier to use the First Fundamental Theorem of Calculus to find $F'(x)$ than what you have tried?

 

[_N3WTON_]

I assume you mean \lim_{x\rightarrow 0} \frac{1}{x} \int_1^{1+x} \frac{\cos t}{t} \, dt This integral has no elementary antiderivative (it is a known special function, the Cosine Integral), so it cannot be solved in a trivial way through the use of the Fundamental Theorem directly on the function as is. Instead, you may want to use the Taylor series for cos(t), as it is an analytic function, and integrate the series that results term by term (be sure to note that it is uniformly convergent on the interval [1, 1+x]). Then multiply by 1/x and take the limit of the series.

Would there be a way to do it without using the Taylor series? Technically in my class we have not learned it yet so I believe it is supposed to be solved without the use of a series.

 

[_N3WTON_]

Given $F(x)=\int_1^{1+x}\frac{\cos t}{t}\ dt$, you have correctly identified that $\lim_{x\rightarrow 0}F(x)=0$ and that l'Hopital's rule might prove useful in evaluating $\lim_{x\rightarrow 0}\frac{F(x)}{x}$.

Might it be easier to use the First Fundamental Theorem of Calculus to find $F'(x)$ than what you have tried?

I can't use the First Theorem because I do not know how to integrate the function

 

[slider142]

I can't use the First Theorem because I do not know how to integrate the function

Recall that the theorem states that "If f is continuous on the interval [a, b], then the function defined by F(x) = \int_a^x f(t) dt is differentiable on (a, b) with derivative F'(x) = f(x)." In other words, you do not need to know an antiderivative of f(t). You just need to know whether it is continuous or not. :)

 

[pasmith]

I can't use the First Theorem because I do not know how to integrate the function

Use the Second Theorem.

 

[_N3WTON_]

Recall that the theorem states that "If f is continuous on the interval [a, x], then the function defined by F(x) = \int_a^x f(t) dt is differentiable on (a, x) with derivative F'(x) = f(x)." In other words, you do not need to know an antiderivative of f(t). You just need to know whether it is continuous or not. :)

So does this mean I could just write the problem all in terms of x and then apply L'Hopsital's theorem? (ie, differentiate the top and the bottom of the equation?)

 

[slider142]

So does this mean I could just write the problem all in terms of x and then apply L'Hopsital's theorem?

Well, to be sure, what are you going to write as the derivative of the numerator \int_1^{1+x} \frac{\cos t}{t}\, dt \; ? It will indeed be a function of x, as the variable t is just a dummy variable of integration.

 

[gopher_p]

I can't use the First Theorem because I do not know how to integrate the function

In most texts that I have worked with, the First Fundamental Theorem essentially states that, for $F(x)=\int_a^xf(t)\ dt$, $F'(x)=f(x)$. I usually tell my students that it's the easiest derivative formula they're ever going to come across. It allows one to find the derivative of a so-called "accumulation function" with almost no effort whatsoever.

A more general version of the First Fundamental Theorem states $$\frac{d}{dx}\left(\int_{u(x)}^{v(x)}f(t)\ dt\right)=f\big(v(x)\big)v'(x)-f\big(u(x)\big)u'(x)$$ whcih is derived from the "normal" First Fundamental Theorem, properties of definite integrals, and the chain rule for derivatives. This is the version that applies here.

 

[_N3WTON_]

Well, to be sure, what are you going to write as the derivative of the numerator \int_1^{1+x} \frac{\cos t}{t}\, dt \; ? It will indeed be a function of x, as the variable t is just a dummy variable of integration.

I was going to write (1+x)(Sin(1+x)) + (Cos(1+x))/ (1+x)^2

 

[_N3WTON_]

After reading about the Second theorem I am thinking it may just be Cos(1+x)/(1+x)

 

[slider142]

I was going to write (1+x)(Sin(1+x)) + (Cos(1+x))/ (1+x)^2

I see. That's not quite correct. Let me apply the First Fundamental Theorem to a different function so you can see how it is used. Suppose we want to know the derivative of g(x) = \int_1^{x + 3} \sin(t)\, dt
Since sin(t) is continuous on any interval [1, b], the First Fundamental Theorem tells us that the function F(x) = \int_1^x \sin(t)\, dt is differentiable on (1, b) with derivative F'(x) = sin(x). Until you gain more experience with applying it, always use the exact wording of the theorem.
Our function is therefore a composition: g(x) = F(x + 3). Therefore, the chain rule tells us that g'(x) = F'(x + 3) * (x + 3)' = sin(x + 3)*1 = sin(x + 3). Does that help ?

 

[slider142]

After reading about the Second theorem I am thinking it may just be Cos(1+x)/(1+x)

There you go! :)

 

[_N3WTON_]

I see. That's not quite correct. Let me apply the First Fundamental Theorem to a different function so you can see how it is used. Suppose we want to know the derivative of g(x) = \int_1^{x + 3} \sin(t)\, dt
Since sin(t) is continuous on any interval [1, b], the First Fundamental Theorem tells us that the function F(x) = \int_1^x \sin(t)\, dt is differentiable on (1, b) with derivative F'(x) = sin(x). Until you gain more experience with applying it, always use the exact wording of the theorem.
Our function is therefore a composition: g(x) = F(x + 3). Therefore, the chain rule tells us that g'(x) = F'(x + 3) * (x + 3)' = sin(x + 3)*1 = sin(x + 3). Does that help ?

It does help, now I just need to apply L'hospital's rule, correct?

 

[_N3WTON_]

Ok, so I don't think I need L'hospital's, applying the limit I get cos(1), correct?===> Nevermind, I don't think this answer is correct

 

[slider142]

yes, but would I not need to use the quotient rule on the derivative?

Remember that we are finding the derivative of g(x) = \int_1^{1+x} \frac{\cos t}{t}\, dt, which is an integral, not a quotient. The First Fundamental Theorem never requires us to find the derivative of the function inside the integral. It tells us that the derivative is the function inside the integral. That is, we are definitively not find the derivative of \frac{\cos t}{t}, as that quantity is not even a function of x! Only the integral varies with x, and the amount that it varies is precisely the value of the function being integrated.
This is fundamentally connected to what we do when we integrate: we add up the values of the function over each interval scaled by the size of each interval as that size approaches 0. So it makes sense that the rate at which the integral varies at each point is the same as the value of the function being added at that point.
Edit: If you meant L'Hopital's rule, then no, you do not differentiate the quotient. As long as the numerator and denominator individually approach 0, you differentiate them as separate functions.

 

[_N3WTON_]

Remember that we are finding the derivative of F(x) = \int_1^{1+x} \frac{\cos t}{t}\, dt, which is an integral, not a quotient. The First Fundamental Theorem never requires us to find the derivative of the function inside the integral. It tells us that the derivative is the function inside the integral. That is, we are definitively not find the derivative of \frac{\cos t}{t}, as that quantity is not even a function of x! Only the integral varies with x, and the amount that it varies is precisely the value of the function being integrated.
This is fundamentally connected to what we do when we integrate: we add up little variations of the function over each interval as the size of each interval approaches 0. So it makes sense that the rate at which the integral varies is the same as the "size" of each piece.

Thank you, so I took the limit as n->0 of ((Cos(1+x)/(1+x))/x and found that the limit does not exist

 

[_N3WTON_]

Thank you, so I took the limit as n->0 of ((Cos(1+x)/(1+x))/x and found that the limit does not exist

However, I simplified the equation and found the limit to be zero, which was not one of the answer provided, so I think I may have made a mistake evaluating the limit

 

[slider142]

Thank you, so I took the limit as n->0 of ((Cos(1+x)/(1+x))/x and found that the limit does not exist

You differentiated the numerator, which was the integral, but you didn't differentiate the denominator of x. Your new limit, due to applying L'Hopital's rule to the original limit, should be
\lim_{x\rightarrow 0} \frac{\int_1^{1+x} \frac{\cos t}{t}\, dt}{x} = \lim_{x\rightarrow 0} \frac{\frac{\cos(1 + x)}{1+x}}{1}
That definitely has a real limit. :)

 

[_N3WTON_]

You differentiated the numerator, which was the integral, but you didn't differentiate the denominator of x. Your new limit, due to applying L'Hopital's rule to the original limit, should be
\lim_{x\rightarrow 0} \frac{\int_1^{1+x} \frac{\cos t}{t}\, dt}{x} = \lim_{x\rightarrow 0} \frac{\frac{\cos(1 + x)}{1+x}}{1}
That definitely has a real limit. :)

I see, thank you very much!