Recent content by goodheavens

i see it now. thank you :)

i've thought of that method also but there's this theorem that we have to use called the theorem on limit of a composite function which states that if lim g(x) as x->a is = b and if the function f is continuous at b, lim (f o g) (x) as x->a is = f(b) or, equivalently, lim f(g(x)) as x->a...

Homework Statement If lim f(x) as x->0 is = 0 then lim \frac{sin(f(x))}{f(x)} as x->0 = 1? dont know how to start proving this . thanks for the replies

oh yeah. haha thank you :smile::

yeah cause its the same as in this https://www.physicsforums.com/showthread.php?t=460116 that would be double posting :) so this is the final proof that i made for the previous question. Let \epsilon > 0 l x-c l < \delta \Rightarrow l f(x) - f(c) l < \epsilon Since f(c) > 0. then...

the interval (a,b) was actually one of the given data. thank you

i forgot to mention that the interval was (a,b)

thank you sir

i saw a similar problem to this one and I am not sure if this works using the \epsilon - \delta definition we have l x-c l < \delta \Rightarrow l f(x) - f(c) l < \epsilon then they used \epsilon = f(c)/2 lf(x)-f(c)l < f(c)/2 -f(c)/2<f(x)-f(c)< f(c)/2 f(c)/2 < f(x) < 3f(c)/2...

how do you prove the sign-preserving property? it says here that. If f is continuous at a, and f(a) < 0, then there is an open interval I containing a such that f(x) < 0 for every x in I. For a proof, simply take the open interval (2f(a),0) for the challenge interval "J" in the...