Limits of frac(x)/x: Is WolframAlpha Wrong?

[SataSata]

$$\lim_{x\to\infty} \frac {frac(x)} {x} $$
frac(x) is x minus floor function of x. So if x = 2.5, floor function = 2 and frac(x) = 0.5
Hence frac(x) will always be a number between -1 and 1 but never -1 and 1.

By squeeze theorem,
-1 < frac(x) < 1
-1/x < frac(x)/x < 1/x
0 < frac(x)/x < 0

Does this means that $$\lim_{x\to\infty} \frac {frac(x)} {x} = Undefined? $$
Since it is between 0 but not 0.
However WolframAlpha gives the answer as 0.
Shouldn't it be 0 only if it is $$0 \leqslant frac(x)/x \leqslant 0$$
So did I do something wrong or is WolframAlpha wrong?

 

[Titan97]

frac(x) is x minus floor function of x. So if x = 2.5, floor function = 2 and frac(x) = 0.5

No. frac{x} is always positive and lies between 0 and 1.

In the limit, the numerator always lies between 0 and 1 while the denominator is an increasing function that tends to infinity when x tends to infinity.

So what will be $\lim_{x\to\infty}\frac{frac(x)}{x}$

 

[SataSata]

Thank you Titan97 for clearing that up. So the limits would be 0 this time.

 

[Mark44]

You can use the squeeze theorem here since $0 \le \text{frac(x)} < 1$.

 

[Titan97]

Thank you Titan97 for clearing that up. So the limits would be 0 this time.

Yes.