If the function f(x) = begin{cases} frac{1}{x | vedclass

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0^{+}} f(x) = f(0) = k$.First,calculate the $RHL$:$\lim

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$-5$

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(A) For $f(x)$ to be continuous at $x = 0$,we must have $\lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0^{+}} f(x) = f(0) = k$.First,calculate the $RHL$:$\lim_{x 	o 0^{+}} f(x) = \lim_{x 	o 0^{+}} \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1} = \lim_{x 	o 0^{+}} \frac{\cos(2x) - 1}{\sqrt{x^{2}+1}-1}$.Using $\cos(2x) - 1 = -2\sin^{2} x$,we get:$\lim_{x 	o 0^{+}} \frac{-2\sin^{2} x}{\sqrt{x^{2}+1}-1} 	imes \frac{\sqrt{x^{2}+1}+1}{\sqrt{x^{2}+1}+1} = \lim_{x 	o 0^{+}} \frac{-2\sin^{2} x (\sqrt{x^{2}+1}+1)}{x^{2}} = -2(1)^{2}(1+1) = -4$.So,$k = -4$.Next,calculate the $LHL$:$\lim_{x 	o 0^{-}} f(x) = \lim_{x 	o 0^{-}} \frac{1}{x} \ln\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}ight) = \lim_{x 	o 0^{-}} \left[ \frac{\ln(1+\frac{x}{a})}{x} - \frac{\ln(1-\frac{x}{b})}{x} ight]$.Using $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,we get:$\frac{1}{a} - (-\frac{1}{b}) = \frac{1}{a} + \frac{1}{b}$.Since $LHL = k$,we have $\frac{1}{a} + \frac{1}{b} = -4$.Finally,calculate $\frac{1}{a} + \frac{1}{b} + \frac{4}{k} = -4 + \frac{4}{-4} = -4 - 1 = -5$.

If the function $f(x) = \begin{cases} \frac{1}{x} \log_{e}\left(\frac{1+\frac{x}{a}}{1-\frac{x}{b}}\right), & x < 0 \\ k, & x = 0 \\ \frac{\cos^{2} x - \sin^{2} x - 1}{\sqrt{x^{2}+1}-1}, & x > 0 \end{cases}$ is continuous at $x = 0$,then $\frac{1}{a} + \frac{1}{b} + \frac{4}{k}$ is equal to:

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