Possibly the hardest limit question i have ever come across.

[mrjoe2]

b]1. Homework Statement [/b]
F(x) = lim csct - cscx find the value of f '(pi/4) ( f prime pi/4)
t-->x t - x

(the denominator is screwed up. it should be t-->x for the limit and t-x for the denominator, but you guys probably already guess that. )

Homework Equations

i don't know. perhaprs first principles. please avoid l'hospital's rule for even i can do that haha. I don't know what I am not seeing, i just can't approach the question.

The Attempt at a Solution

iv tried hundreds of things. the first thing i tried was multiplying top and bottom by conjugates to see if i could work out something with that. the second thing i tried was changing csc into its primitive for 1/sinx and finding a lowest common denominator and using fraction rules. the next thing i tried was separating the top and bottom into functions of their own... that just threw me into a brick wall. please if someone can give me a starting point i would appreciate it soooooo much!

 

[Dick]

Do change csc(x) into 1/sin(x) and rearrange it. I'm guessing you probably know the limit of (sin(t)-sin(x))/(t-x). Use that.

 

[mrjoe2]

Do change csc(x) into 1/sin(x) and rearrange it. I'm guessing you probably know the limit of (sin(t)-sin(x))/(t-x). Use that.

you are completely right, i could find a solution that way if i was to use l'hopital's rule
in the ladder part, after all conversions and reductions have been made. however
i am tending away from his rule because i am in first year calculus and want to
grasp the "primitive" way of completing the question.

 

[Dick]

(sin(t)-sin(x))/(t-x) is a difference quotient for the derivative of the sine function. To do that in a "primitive" way, write it as the limit of (sin(x+h)-sin(x))/h as h->0 and use the trig addition rule. Then use some other "primitive" stuff like lim sin(h)/h=1 as h->0.