Find the limit of 1/x - floor(1/x) as x approaches 0 from the right, where floor denotes the greatest integer function.

Step-by-step solution

1. 1

   Recognise the expression as a fractional part. For any real $t$, $t - \lfloor t \rfloor$ is the fractional part $\{t\}$, which always satisfies $0 \le \{t\} < 1$. So the whole expression is $\left\{\tfrac1x\right\}$ and is trapped in $[0,1)$ — bounded, but boundedness alone never proves a limit exists.

2. 2

   See what the inner function does. As $x \to 0^{+}$, $\tfrac1x \to +\infty$. The fractional part of a quantity racing to infinity sweeps through $[0,1)$ over and over, once per unit increase — and $\tfrac1x$ increases without bound, so it makes infinitely many such sweeps in any interval $(0,\delta)$.

3. 3

   Build a sequence on which the value is always 0. Take $x_n = \tfrac{1}{n}$ for integer $n$. Then $\tfrac{1}{x_n} = n$ is an integer, so

   $\frac{1}{x_n} - \left\lfloor \frac{1}{x_n} \right\rfloor = n - n = 0$

   and $x_n \to 0^{+}$.

4. 4

   Build a second sequence on which the value is always 1/2. Take $y_n = \dfrac{1}{n + \tfrac12}$. Then $\tfrac{1}{y_n} = n + \tfrac12$, whose floor is $n$, so

   $\frac{1}{y_n} - \left\lfloor \frac{1}{y_n} \right\rfloor = \frac12$

   and $y_n \to 0^{+}$ as well.

5. 5

   Conclude by the sequential criterion. If the limit existed and equalled $L$, every sequence approaching $0^{+}$ would force the values to $L$. Two sequences give $0$ and $\tfrac12$, so no such $L$ exists:

   $\lim_{x \to 0^{+}} \left(\frac{1}{x} - \left\lfloor \frac{1}{x} \right\rfloor\right) \text{ does not exist}$

   In fact by choosing $x = \tfrac{1}{n + c}$ for any $c \in [0,1)$ the value can be made exactly $c$, so every number in $[0,1)$ is a subsequential limit.