Question

Let $f : \mathbb{R} \to \mathbb{R}$ be a function defined by \[ f(x) = \begin{cases} x \left( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \right) & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \] Then

• A. there exists a constant C such that $|f(x)| \le C$ for all $x \in \mathbb{R}$.
• B. f is monotonically increasing in the interval (-1, 1).
• C. f is differentiable at $x = 0$.
• D. f is not continuous at $x = 0$.

Hint:

For any function of the form $x \cdot g(x)$ where $g(x)$ is bounded, the limit as $x \to 0$ is always 0. This quickly guarantees continuity at 0.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem presents a piecewise-defined function and asks us to evaluate its continuity, differentiability, monotonicity, and boundedness.

Step 2: Key Formula or Approach:
We can write the function in terms of hyperbolic tangent:
\[ f(x) = x \tanh\left(\frac{1}{x}\right) \text{ for } x \neq 0 \]
We will use standard limits and inequalities for $\tanh(u)$.

Step 3: Detailed Explanation:

• Let us analyze continuity at $x = 0$:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} x \tanh\left(\frac{1}{x}\right) \]
Since $|\tanh(u)| \lt 1$ for all real numbers $u$, we have:
\[ |f(x)| = |x| \left|\tanh\left(\frac{1}{x}\right)\right| \le |x| \]
As $x \to 0$, $|x| \to 0$. By the Squeeze Theorem, $\lim_{x \to 0} f(x) = 0 = f(0)$.
Therefore, $f(x)$ is continuous at $x = 0$. This makes option (D) false.

• Let us check differentiability at $x = 0$:
Using the definition of the derivative:
\[ f'(0) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = \lim_{h \to 0} \frac{h \tanh(1/h) - 0}{h} = \lim_{h \to 0} \tanh\left(\frac{1}{h}\right) \]
- As $h \to 0^+$, $\frac{1}{h} \to \infty$, so $\tanh(1/h) \to 1$.
- As $h \to 0^-$, $\frac{1}{h} \to -\infty$, so $\tanh(1/h) \to -1$.
Since the left-hand and right-hand limits are different, $f(x)$ is not differentiable at $x = 0$. This makes option (C) false.

• Let us check monotonicity:
The function $f(x) = x \tanh(1/x)$ is an even function because:
\[ f(-x) = (-x) \tanh(-1/x) = (-x)(-\tanh(1/x)) = x \tanh(1/x) = f(x) \]
A non-constant even function cannot be monotonically increasing on a symmetric interval around 0. This makes option (B) false.

• Now let us check boundedness:
For any $x \neq 0$, let $u = \frac{1}{|x|} \gt 0$. Then:
\[ |f(x)| = |x| \tanh\left(\frac{1}{|x|}\right) = \frac{\tanh(u)}{u} \]
Since $\tanh(u) \lt u$ for all $u \gt 0$, we have:
\[ \frac{\tanh(u)}{u} \lt 1 \implies |f(x)| \lt 1 \text{ for all } x \in \mathbb{R} \]
Thus, the function is bounded by $C = 1$. This makes option (A) correct.



Step 4: Final Answer:
There exists a constant $C$ such that $|f(x)| \le C$ for all $x \in \mathbb{R}$.