If the function f defined as f(x) = frac{1}{x | vedclass

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(A) If the function is continuous at $x = 0$,then $\lim_{x 	o 0} f(x)$ must exist and $f(0) = \lim_{x 	o 0} f(x)$.Now,$\lim_{x 	o 0} f(x) = \lim_{x

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$(3, 1)$

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(A) If the function is continuous at $x = 0$,then $\lim_{x 	o 0} f(x)$ must exist and $f(0) = \lim_{x 	o 0} f(x)$.Now,$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \left( \frac{1}{x} - \frac{k - 1}{e^{2x} - 1} ight)$.$= \lim_{x 	o 0} \left( \frac{e^{2x} - 1 - (k - 1)x}{x(e^{2x} - 1)} ight)$.Using the expansion $e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots$,we get:$= \lim_{x 	o 0} \frac{(1 + 2x + 2x^2 + \dots) - 1 - kx + x}{x(2x + 2x^2 + \dots)}$.$= \lim_{x 	o 0} \frac{(3 - k)x + 2x^2 + \dots}{2x^2 + 2x^3 + \dots}$.For the limit to exist,the coefficient of $x$ in the numerator must be zero,so $3 - k = 0$,which gives $k = 3$.Substituting $k = 3$,the limit becomes $\lim_{x 	o 0} \frac{2x^2 + \dots}{2x^2 + 2x^3 + \dots} = \frac{2}{2} = 1$.Therefore,$f(0) = 1$,and the ordered pair is $(3, 1)$.

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 the ordered pair $(k, f(0))$ is equal to?

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