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: Rewrite the function compactly.
For $x \ne 0$, $\frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} = \tanh\!\left(\frac{1}{x}\right)$. So $f(x) = x \tanh\!\left(\frac{1}{x}\right)$, and $f(0) = 0$.

Step 2: Note the size of $\tanh$.
For every real input, $\tanh$ stays strictly between $-1$ and $1$, so $\left|\tanh\!\left(\frac{1}{x}\right)\right| \lt 1$.

Step 3: Bound the function by $|x|$.
Therefore $|f(x)| = |x|\left|\tanh\!\left(\frac{1}{x}\right)\right| \lt |x|$. We test whether this forces a global bound.

Step 4: Check large $x$.
As $x$ grows large, $\frac{1}{x}$ is tiny and $\tanh\!\left(\frac{1}{x}\right) \approx \frac{1}{x}$, so $f(x) \approx x \cdot \frac{1}{x} = 1$. The function approaches $1$ but its absolute value never reaches or exceeds $1$.

Step 5: Confirm a constant bound exists.
Combining: for small $|x|$ we have $|f(x)| \lt |x|$ which is small, and for large $|x|$ the value sits below $1$. Across all of $\mathbb{R}$, $|f(x)| \lt 1$. So we may take $C = 1$, and $|f(x)| \le C$ holds for every $x$.

Step 6: Reject the other options.
The function is bounded, so the statement that a bounding constant exists is the correct one.
\[ \boxed{\text{there exists a constant } C \text{ with } |f(x)| \le C \text{ for all } x} \]