Question

The function $f:R/0\to R$ given by $f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1}$ can be made continuous at $x=0$ by
defining $f\left(0\right)$ as

• A. 0
• B. 1
• C. 2
• D. -1

Answer 1

Answer: (B)

1

Given $f\left(x\right)=\frac{1}{x}-\frac{2}{e^{2x}-1}$

For $f\left(x\right)$ to be continuous at $x=0$ , its limit should exists at $x=0$ and must be finite.

Hence

$f\left(0\right)=\underset{x\to 0}{lim}\frac{e^{2x}-1-2x}{x(e^{2x}-1)}$

$=\underset{x\to 0}{lim}\frac{1+2x+\frac{4x^2}{2!}+\frac{8x^3}{3!}.....-1-2x}{x+2x^2+\frac{4x^3}{2!}.....-x}=\frac{\frac{4x^2}{2!}}{2x^2}=1$......[neglecting higher power of $x$]