If f(x) = frac{e^{2x} - (1 + 4x)^{1/2}}{ln(1 | vedclass

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(B) To find the limit of $f(x)$ as $x 	o 0$,we use the Taylor series expansions:$e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1

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a removable discontinuity at $x = 0$ and $f(0) = -4$

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(B) To find the limit of $f(x)$ as $x 	o 0$,we use the Taylor series expansions:$e^{2x} = 1 + 2x + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + 2x^2 + \frac{4x^3}{3} + \dots$$(1 + 4x)^{1/2} = 1 + \frac{1}{2}(4x) + \frac{\frac{1}{2}(\frac{1}{2}-1)}{2!}(4x)^2 + \dots = 1 + 2x - 2x^2 + \dots$$\ln(1 - x^2) = -x^2 - \frac{(-x^2)^2}{2} - \dots = -x^2 - \frac{x^4}{2} - \dots$Substituting these into the expression for $f(x)$:$f(x) = \frac{(1 + 2x + 2x^2 + \dots) - (1 + 2x - 2x^2 + \dots)}{-x^2 - \frac{x^4}{2} - \dots}$$f(x) = \frac{4x^2 + \dots}{-x^2 - \dots}$Taking the limit as $x 	o 0$:$\lim_{x 	o 0} f(x) = \lim_{x 	o 0} \frac{4x^2}{-x^2} = -4$Since the limit exists and is finite,$f$ has a removable discontinuity at $x = 0$ and $f(0) = -4$.

$A$. $\|x\| + \|x - 2\|$	$I$. Right hand limit does not exist at $x = 2$.
$B$. $\text{cosech } x$	$II$. Continuous only for non-zero real values of $x$.
$C$. $x - [x]$	$III$. Limit is zero for all real $x$.
$D$. $\sqrt{2 - x}$	$IV$. Continuous for all real value of $x$.
	$V$. Discontinuous at all integral values of $x$.

If $f(x) = \frac{e^{2x} - (1 + 4x)^{1/2}}{\ln(1 - x^2)}$ for $x \neq 0$,then $f$ has

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