Understanding the Limit Problem

The question asks us to find the limit of the function $f(x)$ as $x$ approaches 0. The function $f(x)$ is defined differently based on the value of the greatest integer function, $[x]$.

The function is given as:

$$ f(x) = \begin{cases} \dfrac{\sin [x]}{[x]}, & \text{if } [x] \neq 0 \\ 0, & \text{if } [x] = 0 \end{cases} $$

Here, $[x]$ denotes the greatest integer function, which returns the largest integer less than or equal to $x$. For example, $[3.7] = 3$ and $[-2.3] = -3$.

Analyzing the Greatest Integer Function Near Zero

To find the limit as $x \rightarrow 0$, we need to examine the behavior of the function as $x$ approaches 0 from both the left side ($x \rightarrow 0^-$) and the right side ($x \rightarrow 0^+$).

Right-Hand Limit (x → 0⁺)

When $x$ approaches 0 from the right side, $x$ takes small positive values (e.g., 0.1, 0.01, 0.001). For any $x$ such that $0 < x < 1$, the greatest integer function $[x]$ evaluates to 0.

According to the function's definition, when $[x] = 0$, $f(x) = 0$.

Therefore, the right-hand limit is:

$$ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 0 = 0 $$

Left-Hand Limit (x → 0^-)

When $x$ approaches 0 from the left side, $x$ takes small negative values (e.g., -0.1, -0.01, -0.001). For any $x$ such that $-1 < x < 0$, the greatest integer function $[x]$ evaluates to -1.

In this case, $[x] = -1$, which is not equal to 0. So, we use the first part of the function definition: $f(x) = \frac{\sin[x]}{[x]}$.

Substituting $[x] = -1$ into the function:

$$ f(x) = \frac{\sin(-1)}{-1} $$

Since $\sin(-y) = -\sin(y)$, we have:

$$ f(x) = \frac{-\sin(1)}{-1} = \sin(1) $$

Therefore, the left-hand limit is:

$$ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \sin(1) = \sin(1) $$

Comparing the Limits

For the overall limit $\lim_{x \to 0} f(x)$ to exist, the left-hand limit and the right-hand limit must be equal.

• Right-Hand Limit: $\lim_{x \to 0^+} f(x) = 0$
• Left-Hand Limit: $\lim_{x \to 0^-} f(x) = \sin(1)$

We know that $\sin(1)$ is approximately 0.8415. Since $0 \neq \sin(1)$, the left-hand limit and the right-hand limit are not equal.

Conclusion

Because the limit from the left does not equal the limit from the right, the limit of the function $f(x)$ as $x$ approaches 0 does not exist.