Is the function f(x)=sinxxf(x) = \frac{\sin x}{x}f(x)=xsinx for x≠0x \neq 0x=0 and f(0)=1f(0) = 1f(0)=1 continuous at x=0x=0x=0?
Yes, because limx→0sinxx=1=f(0)\lim_{x \to 0} \frac{\sin x}{x} = 1 = f(0)limx→0xsinx=1=f(0)
No, because sin(0)0\frac{\sin(0)}{0}0sin(0) is undefined
No, because the limit does not exist
Yes, but only from the right