The value of operatorname{Lt}_{x rightarrow 0 | vedclass

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(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for t

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$e^2$

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(A) We use the standard limit formula $\operatorname{Lt}_{x \rightarrow a} [f(x)]^{g(x)} = e^{\operatorname{Lt}_{x \rightarrow a} g(x)[f(x)-1]}$ for the indeterminate form $1^{\infty}$.Here,$f(x) = \frac{1+5x^2}{1+3x^2}$ and $g(x) = \frac{1}{x^2}$.As $x \rightarrow 0$,$f(x) \rightarrow 1$ and $g(x) \rightarrow \infty$.Thus,the limit is $e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2}{1+3x^2} - 1 \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{1+5x^2 - (1+3x^2)}{1+3x^2} \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{1}{x^2} \left( \frac{2x^2}{1+3x^2} \right)}$.$= e^{\operatorname{Lt}_{x \rightarrow 0} \frac{2}{1+3x^2}}$.$= e^{\frac{2}{1+0}} = e^2$.

The value of $\operatorname{Lt}_{x \rightarrow 0} \left( \frac{1+5x^2}{1+3x^2} \right)^{\frac{1}{x^2}}$ is

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