Question

How would you solve for the lim x-> 0 of (integral 1 to 1+x of (cos t / t) dt)/x ?

Answer 1

\[\lim_{x \rightarrow 0} \int_{1}^{1+x} \frac{\cos t}{t} dt\]
right??

Answer 2

The integral is the numerator with x as a denominator*

Answer 3

uhh complicated..i'll call the os genius @satellite73 :)

Answer 4

he'll LOVE that hehe

Answer 5

i think you can use l'hopital

Answer 6

I would do that, but I don't know how to get the derivative of the integral in this case

Answer 7

\[f(x)=\int_1^{1+x}\frac{\cos(t)}{t}dt\]
\[g(x)=x\] and you want
\[\lim_{x\to 0}\frac{f(x)}{g(x)}\]

Answer 8

derivative of the integral is the integrand so you get
\[f'(x)=\frac{\cos(1+x)}{x}\],
\[g'(x)=1\]

Answer 9

now take the limit as x goes to zero, and hmmm something is wrong

Answer 10

oh my derivative is wrong!

Answer 11

I know the answer is cos 1, if that helps. I just don't understand why/how.

Answer 12

\[f'(x)=\frac{\cos(1+x)}{1+x}\]

Answer 13

i forgot to replace the \(t\) in the integrand by \(1+x\) in the denominator as well as in the numerator
now take
\[\lim_{x\to 0}\frac{\cos(1+x)}{1+x}=\cos(1)\]

Answer 14

does the lower bound of the integral become irrelevant in this type of problem?

Answer 15

only part that requires thinking is taking the derivative of
\[f(x)=\int_1^{1+x}\frac{\cos(t)}{t}dt\] using the chain rule

Answer 16

you mean for taking the derivative? yes, the lower bound is irrelevant if it is a number and does not include a function of \(x\)

Answer 17

*if it does not include a function of \(x\)

Answer 18

the lower bound determines the constant, but the derivative of a constant is 0

Answer 19

makes sense, thank you!

Answer 20

yw