mathop {lim }limits_{x to 0} {left( {frac{{1 | vedclass

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Question

(A) We use the standard limit formula: $\mathop {\lim }\limits_{x 	o 0} (1 + f(x))^{1/g(x)} = e^{\mathop {\lim }\limits_{x 	o 0} \frac{f(x)}{g(x)}}$

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$e^2$

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(A) We use the standard limit formula: $\mathop {\lim }\limits_{x 	o 0} (1 + f(x))^{1/g(x)} = e^{\mathop {\lim }\limits_{x 	o 0} \frac{f(x)}{g(x)}}$.Given the expression $\mathop {\lim }\limits_{x 	o 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} ight)^{1/{x^2}}}$,we can rewrite the base as $1 + \left( \frac{1 + 5x^2}{1 + 3x^2} - 1 ight) = 1 + \frac{2x^2}{1 + 3x^2}$.Thus,the limit becomes $e^{\mathop {\lim }\limits_{x 	o 0} \frac{1}{x^2} \cdot \frac{2x^2}{1 + 3x^2}}$.Simplifying the exponent: $\mathop {\lim }\limits_{x 	o 0} \frac{2}{1 + 3x^2} = \frac{2}{1 + 0} = 2$.Therefore,the result is $e^2$.

$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = $

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