Search results

good point! the limit exists by the ε - δ definition but the graph has a hole at x = 2 and is therefore not continuous

but for the 0⁰ case the lhs limit cannot be looked at because the gradient for x^x is x^x(log_ex+1) which is not real if x <= 0 so no real gradient exists for the graph of x^x for x <= 0 and hence the graph does not exist for x <= 0

im sorry, but i dont understand what u mean: i can use l'hopital's rule because: also, many sources agree that 0⁰ = 1 Google: Mathematica: Maple:

wow, just read FinalFantasy's solution, its really neat & creative, id have never seen that way FinalFantasy is by far the best at integration on this forum, ive noticed he seems very modest about it though by the way who_loves_maths thanks for defending me before on the 0⁰...

proved u can use l'hopital's rule (look at my last posts) i did have a logical basis, all i did wrong was say x → 0 instead of x → 0⁺ this is just a notation error, my argument for 0⁰ was completely correct maybe u should stop wasting our time with ur stupid...

hey acmilan, this is an interesting thread but should it be moved to the Appreciating the Beauty and Elegance (extracurricular topics) forum? thanks also, i just want to take back my comment before when i said u werent an excellent mod. i read over more of ur posts and u were extremely helpful

good point! 2u students often forget that they must use radians when doing anything that involves calculus this is because in the derivation of the gradient of sin x, the gradient of sin x only equals cos x when: which is only true when x is in radians, put ur calc into radians and...

yes u can! to evaluate the limit of xlog_ex as x → 0 using l'hopitals rule: [EDIT: this should be x → 0⁺, just so Xayma doesnt have a cry lol] just for those interested, heres a proof of l'hopitals rule the gradient of x^x is...