If f(x) = frac{ln(e^{x^2} + 2sqrt{x})}{sqrt{x | vedclass

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(D) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0^+} f(x)$.$f(0) = \lim_{x 	o 0^+} \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{

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$2$

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(D) For $f(x)$ to be continuous at $x = 0$,we must have $f(0) = \lim_{x 	o 0^+} f(x)$.$f(0) = \lim_{x 	o 0^+} \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$Using the standard limit $\lim_{u 	o 0} \frac{\ln(1+u)}{u} = 1$,we rewrite the expression as:$f(0) = \lim_{x 	o 0^+} \frac{\ln(1 + (e^{x^2} + 2\sqrt{x} - 1))}{e^{x^2} + 2\sqrt{x} - 1} 	imes \frac{e^{x^2} + 2\sqrt{x} - 1}{\sqrt{x}}$As $x 	o 0^+$,$e^{x^2} + 2\sqrt{x} - 1 	o 0$,so the first part of the limit is $1$.$f(0) = 1 	imes \lim_{x 	o 0^+} \left( \frac{e^{x^2} - 1}{\sqrt{x}} + \frac{2\sqrt{x}}{\sqrt{x}} ight)$$f(0) = \lim_{x 	o 0^+} \left( \frac{e^{x^2} - 1}{x^2} \cdot x^{3/2} + 2 ight)$Since $\lim_{x 	o 0^+} \frac{e^{x^2} - 1}{x^2} = 1$ and $\lim_{x 	o 0^+} x^{3/2} = 0$,the first term vanishes.$f(0) = 0 + 2 = 2$.

If $f(x) = \frac{\ln(e^{x^2} + 2\sqrt{x})}{\sqrt{x}}$ is continuous at $x = 0$,then $f(0)$ must be equal to:

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