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(A) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1}$,we divide the numerator and the denominator by $e^{x^4}$.$\lim _{

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(A) To evaluate the limit $\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1}$,we divide the numerator and the denominator by $e^{x^4}$.$\lim _{x \rightarrow \infty} \frac{e^{x^4}(1 - \frac{1}{e^{x^4}})}{e^{x^4}(1 + \frac{1}{e^{x^4}})} = \lim _{x \rightarrow \infty} \frac{1 - \frac{1}{e^{x^4}}}{1 + \frac{1}{e^{x^4}}}$.As $x \rightarrow \infty$,$x^4 \rightarrow \infty$,so $e^{x^4} \rightarrow \infty$.Therefore,$\frac{1}{e^{x^4}} \rightarrow 0$.Substituting this into the expression,we get $\frac{1 - 0}{1 + 0} = 1$.

$\lim _{x \rightarrow \infty} \frac{e^{x^4}-1}{e^{x^4}+1} = $

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