Recent content by Jbjohnson15

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

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)...

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.

Ok, so after applying l'hospital's 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.

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?

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

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.

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