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
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...
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...