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

JEE Main · 2007 · 29 Apr 2007 · Offline · Continuity and Differentiability

The function

f : R \sim {0} \to R

given by

f (x) = \frac{1}{x} - \frac{2}{e ^{2 x} - 1}

can be made continuous at

x = 0

by defining

f (0)

Answer: (D)

For continuity at

x = 0

we need

f (0) = lim_{x \to 0} (\frac{1}{x} - \frac{2}{e ^{2 x} - 1})

. Combining over a common denominator gives

lim_{x \to 0} \frac{e ^{2 x} - 1 - 2 x}{x ( e ^{2 x} - 1 )}

, a

\frac{0}{0}

form. Expanding

e^{2 x} = 1 + 2 x + 2 x^{2} + \dots

makes the numerator behave as

2 x^{2}

and the denominator as

2 x^{2}

, whose ratio tends to

1

. Defining

f (0) = 1

therefore makes the function continuous.

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option