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If $\lim _{x \rightarrow 0} \frac{\left(1+a^3\right)+8 e^{1 / x}}{1+\left(1-b^3\right) e^{1 / x}}=2$, then
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Verified Answer
The correct answer is:
$a=1, b=-3^{1 / 3}$
Divide and multiply by $e^{1 / x}$, then
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(1+a^3\right) e^{-1 / x}+8}{e^{-1 / x}+\left(1-b^3\right)}=2 \\
\Rightarrow \quad & \frac{0+8}{0+1-b^3}=2 \Rightarrow 1-b^3=4 \Rightarrow b^3=-3 \\
\Rightarrow \quad & b=-3^{1 / 3}
\end{aligned}
$$
Then, $a \in R$.
$$
\begin{aligned}
& \lim _{x \rightarrow 0} \frac{\left(1+a^3\right) e^{-1 / x}+8}{e^{-1 / x}+\left(1-b^3\right)}=2 \\
\Rightarrow \quad & \frac{0+8}{0+1-b^3}=2 \Rightarrow 1-b^3=4 \Rightarrow b^3=-3 \\
\Rightarrow \quad & b=-3^{1 / 3}
\end{aligned}
$$
Then, $a \in R$.
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