F.6 maths

(a)
f(x) is continuous at x=0
Proof
Since -1<=sin(1/x)<=1
-|x|<=xsin(1/x)<=|x|
lim(x-> 0) -|x|<=lim(x-> 0) xsin(1/x)<=lim(x-> 0) |x|
0<=lim(x-> 0) xsin(1/x)<=0
So lim(x-> 0) xsin(1/x)=0
(b)
f(x) dose not have a derivative at x=0
Proof:
Consider the limit lim[x -> 0] [(F(x) - F(0)) / (x - 0)]. If this limit exist, then F is differentiable at x = 0.
Note that
(F(x) - F(0)) / (x - 0)
= [xsin(1 / x) - 0] / x
= sin(1 / x)
However, lim[x -> 0] sin(1 / x) does not exist.
Therefore F is NOT differentiable at x = 0.
Note:
1 Prove that lim[x -> 0] sin(1 / x) does not exist.
Suppose the limit did exist, then there would be an L such that given an ε＞0,then |x| ＜ σ such that |sin(1/x)-L| ＜ε

let x1=1/(Nπ) such that |x1| ＜ σ

|sin(1/x1)-L|=|L|＜ε
now let x2=1/[(2n 1/2)π] such that |x2| ＜ σ

|sin(1/x2)-L|=|1-L|＜ε

but if take ε＜1/2, L will be close to both 0 and 1,a contradiction.

lim(x→0)(sin1/x) does not exist
2
From the graph of f(x), we can see that when the point tends to 0, the
secant line y oscillatiates between y=x and y=-x, so f(x) cannot differentiate