mathop {lim }limits_{x to 0} frac{d}{{dx}}lef | vedclass

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(C) Let $f(x) = \frac{e^{e^{x^2}} - e}{x}$. We need to find $\lim_{x 	o 0} f'(x)$.Using the quotient rule,$f'(x) = \frac{x \cdot \frac{d}{dx}(e^{e^{x

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$e$

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(C) Let $f(x) = \frac{e^{e^{x^2}} - e}{x}$. We need to find $\lim_{x 	o 0} f'(x)$.Using the quotient rule,$f'(x) = \frac{x \cdot \frac{d}{dx}(e^{e^{x^2}} - e) - (e^{e^{x^2}} - e) \cdot 1}{x^2}$.$f'(x) = \frac{x \cdot (e^{e^{x^2}} \cdot e^{x^2} \cdot 2x) - (e^{e^{x^2}} - e)}{x^2}$.$f'(x) = 2e^{e^{x^2}} \cdot e^{x^2} - \frac{e^{e^{x^2}} - e}{x^2}$.As $x 	o 0$,$2e^{e^{x^2}} \cdot e^{x^2} 	o 2e^1 \cdot e^0 = 2e$.For the second term,let $u = x^2$. As $x 	o 0$,$u 	o 0$. The limit becomes $\lim_{u 	o 0} \frac{e^{e^u} - e}{u} = \lim_{u 	o 0} e \cdot \frac{e^{e^u - 1} - 1}{u}$.Using the standard limit $\lim_{t 	o 0} \frac{e^t - 1}{t} = 1$,where $t = e^u - 1$,we get $e \cdot \lim_{u 	o 0} \frac{e^u - 1}{u} = e \cdot 1 = e$.Thus,the limit is $2e - e = e$.

$\mathop {\lim }\limits_{x \to 0} \frac{d}{{dx}}\left( {\frac{{{e^{{e^{{x^2}}}}} - e}}{x}} \right)$ is

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