Continuity of x*sin(1/x) at x=0

[Miike012]

my book says, the function y = x*sin(1/x) is not continuous at x = 0, however by defining a new function by



F(x) =
x*sin(1/x) , x ≠ 0
0 , x = 0

then F is continuous at x = 0.

This does not make sense to me because the limit as x → 0 is equal to 1, not zero, so therefore there would be a jump discountinuity at x = 0.

unless my calculus is wrong or my understanding is wrong... but isn't
Lim [x*sin(1/x)] = Lim [(x/x)*sin(1/x)/(1/x)] = (1)*(1) = 1 as x approaches zero.
x → 0

 

[CompuChip]

I think your calculus is wrong, unfortunately

For example, using the squeeze theorem: -x \le x \sin(1/x) \le x, therefore
\lim_{x \to 0} -x \le \lim_{x \to 0} \le x \sin(1/x) \le \lim_{x \to 0} x
hence
0 \le \lim_{x \to 0} \le x \sin(1/x) \le 0 \implies \lim_{x \to 0} x \sin(1/x) = 0

You are probably confused with
\lim_{x \to 0} \frac{1}{x} \sin(x) = 1
but if you replace x -> 1/x what you have is the "opposite":
\lim_{x \to \infty} \frac{1}{x} \sin(x) = 1

 

[Miike012]

Im not sure what I am reading towards the end...

I think it says,

Lim as x approaches 0 of x*sin(1/x),
If we let u = 1/x then as x approaches zero, u approaches infinity, so

Lim as u appraoches infinity of sin(u)/u is eqaul to zero... is this correct?

 

[CompuChip]

Sorry about that, I made a typo.
Of course I meant
\lim_{x \to 0} \frac{1}{x} \sin x = 1
instead of 0 (corrected in the above post now).

And yes, I switched to u = 1/x which means x \to 0 changes to u \to \infty (and you replace x with 1/u everywhere).