Q.$\lim_{x \to 3^+} \dfrac{x}{[x]} =$ ________.

The key idea is that for a limit involving the greatest integer function $[x]$, we must consider the behaviour of $[x]$ as $x$ approaches 3 from the right.

For $x \to 3^+$, $x$ is slightly greater than 3, so $[x] = 3$. The function becomes $\dfrac{x}{3}$.

Taking the limit:

The key here is understanding what $[x]$ means — the greatest integer less than or equal to $x$, also called the floor function. For any $x$ just greater than 3, say $x = 3.1$, $[x] = 3$. This is constant on the interval $(3, 4)$, so the denominator behaves like a fixed number near $x = 3^+$.

Because the numerator $x$ approaches 3, and the denominator is exactly 3 for all $x$ in a right-neighbourhood of 3, the limit is straightforward — no factoring, no cancellation, just direct substitution into a constant denominator.

1. Identify the behaviour of $[x]$ near $x = 3^+$.

   For $x$ in $(3, 4)$, the greatest integer less than or equal to $x$ is 3. So $[x] = 3$ for all $x$ with $3 < x < 4$.

2. Rewrite the function for $x$ just to the right of 3.

   Since $[x] = 3$ in that region, we have

3. Take the limit as $x \to 3^+$. Now the limit becomes $$$$