Question

\(f:\mathbb{R}-\{0\}\rightarrow\mathbb{R}\) given by \[ f(x)=\frac{1}{x}-\frac{2}{e^{2x}-1} \] can be made continuous at \(x=0\) by defining \(f(0)\) as:

• A. \(1\)
• B. \(2\)
• C. \(-1\)
• D. \(0\)

Hint:

To make a function continuous at a point: \[ f(a)=\lim_{x\to a} f(x) \] For exponential limits, use the expansion \(e^x-1 \approx x\) as \(x \to 0\).

Answer 1

The Correct Option is A

Step 1: Condition for continuity at \(x=0\). For \(f(x)\) to be continuous at \(x=0\), \[ f(0)=\lim_{x\to 0} f(x) \]
Step 2: Evaluate the limit. \[ \lim_{x\to 0}\left(\frac{1}{x}-\frac{2}{e^{2x}-1}\right) \] Use the standard expansion: \[ e^{2x}-1 = 2x + 2x^2 + \cdots \] \[ \frac{2}{e^{2x}-1} = \frac{2}{2x(1+x+\cdots)} = \frac{1}{x}(1-x+\cdots) \]
Step 3: Substitute into the expression. \[ \frac{1}{x}-\left(\frac{1}{x}-1+\cdots\right) = 1 \]
Step 4: Define \(f(0)\). \[ f(0)=1 \]
Hence, the function can be made continuous at \(x=0\) by defining \[ \boxed{f(0)=1} \]