If the function f defined as $$f(x) = \frac{1}{x} - \frac{k-1}{e^{2x} - 1}$$, $$x \neq 0$$ is continuous at $$x = 0$$, then ordered pair (k, f(0)) is equal to: 

We want the function

$$f(x)=\frac{1}{x}-\frac{k-1}{e^{2x}-1},\qquad x\neq 0$$

to be continuous at $$x=0$$. By the definition of continuity, we must have

$$\lim_{x\to 0}f(x)=f(0).$$

So we first compute the limit as $$x\to 0$$ and then choose $$k$$ so that this limit exists and is finite.

We note that, near zero, each of the two terms $$\dfrac{1}{x}$$ and $$\dfrac{k-1}{e^{2x}-1}$$ blows up, but it is possible that the infinities cancel each other. To see the cancellation clearly, we expand the exponential in a power series.

The Maclaurin series for the exponential function is

$$e^{t}=1+t+\frac{t^{2}}{2!}+\frac{t^{3}}{3!}+\cdots.$$

Taking $$t=2x$$ we have

$$e^{2x}=1+2x+\frac{(2x)^{2}}{2}+\frac{(2x)^{3}}{6}+\cdots =1+2x+2x^{2}+\frac{4}{3}x^{3}+\cdots.$$

Subtracting $$1$$ from both sides gives

$$e^{2x}-1=2x+2x^{2}+\frac{4}{3}x^{3}+\cdots.$$

Now we study the troublesome fraction

$$\frac{k-1}{e^{2x}-1}.$$

First take out the factor $$2x$$ from the denominator:

$$e^{2x}-1=2x\left(1+x+\frac{2}{3}x^{2}+\cdots\right).$$

Hence

$$\frac{k-1}{e^{2x}-1} =\frac{k-1}{2x}\;\frac{1}{\,1+x+\frac{2}{3}x^{2}+\cdots\,}.$$

We now use the standard expansion

$$\frac{1}{1+z}=1-z+z^{2}-\cdots$$

valid for small $$z$$. Here $$z=x+\dfrac{2}{3}x^{2}+\cdots$$, so keeping only the terms up to order $$x$$ gives

$$\frac{1}{1+x+\frac{2}{3}x^{2}+\cdots}=1-x+\cdots.$$

Thus

$$\frac{k-1}{e^{2x}-1} =\frac{k-1}{2x}\Bigl(1-x+\cdots\Bigr) =\frac{k-1}{2x}-\frac{k-1}{2}+\cdots.$$

Now substitute this approximation into $$f(x)$$:

$$\begin{aligned} f(x) &=\frac{1}{x}-\left(\frac{k-1}{2x}-\frac{k-1}{2}+\cdots\right)\\ &=\left(\frac{1}{x}-\frac{k-1}{2x}\right)+\frac{k-1}{2}+\cdots.\\ \end{aligned}$$

We see a term of the form $$\dfrac{\text{something}}{x}$$. For the limit as $$x\to 0$$ to be finite, the coefficient of $$\dfrac{1}{x}$$ must be zero. Therefore we impose

$$1-\frac{k-1}{2}=0.$$

Solving this simple linear equation:

$$1=\frac{k-1}{2}\;\;\Longrightarrow\;\;k-1=2\;\;\Longrightarrow\;\;k=3.$$

With $$k=3$$ the unwanted $$\dfrac{1}{x}$$ term disappears, and we can read off the constant term that remains:

$$f(x)=\cancel{\left(\frac{1}{x}-\frac{2}{2x}\right)}+\frac{2}{2}+\cdots =1+\cdots.$$

Hence

$$\lim_{x\to 0}f(x)=1.$$

To make the function continuous we define

$$f(0)=1.$$

Thus the ordered pair $$(k,f(0))$$ that works is

$$(3,1).$$

Hence, the correct answer is Option B.