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If $\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2$, then
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Verified Answer
The correct answer is:
$a=1, b \in R$
$\lim _{x \rightarrow \infty}\left(1+\frac{a}{x}+\frac{b}{x^2}\right)^{2 x}=e^2$
$$
\Rightarrow \lim _{x \rightarrow \infty}\left(1+\frac{(a x+b)}{x^2}\right)^{2 x}=e^2
$$
$\lim _{x \rightarrow \infty} \frac{a x+b}{x^2}$ must be equal to zero.
$$
\Rightarrow \quad \lim _{x \rightarrow \infty} \frac{a x+b}{x^2}=0
$$
$$
\begin{aligned}
& \Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{a}{x}+\frac{b}{x^2}\right)=0 \\
& \therefore a \text { and } b \in R \\
& \text { From Eq. (i), } \lim _{e^{x \rightarrow \infty}}\left(1+\frac{a x+b}{x^2}-1\right) 2 x=e^2 \\
& \Rightarrow \quad \lim _{c^{x \rightarrow \infty}} \frac{2(a x+b)}{x}=e^2 \\
& \therefore \quad e^{2 a}=e^2 \\
& \Rightarrow \quad a=1 \text { and } b \in R
\end{aligned}
$$
$$
\Rightarrow \lim _{x \rightarrow \infty}\left(1+\frac{(a x+b)}{x^2}\right)^{2 x}=e^2
$$
$\lim _{x \rightarrow \infty} \frac{a x+b}{x^2}$ must be equal to zero.
$$
\Rightarrow \quad \lim _{x \rightarrow \infty} \frac{a x+b}{x^2}=0
$$
$$
\begin{aligned}
& \Rightarrow \quad \lim _{x \rightarrow \infty}\left(\frac{a}{x}+\frac{b}{x^2}\right)=0 \\
& \therefore a \text { and } b \in R \\
& \text { From Eq. (i), } \lim _{e^{x \rightarrow \infty}}\left(1+\frac{a x+b}{x^2}-1\right) 2 x=e^2 \\
& \Rightarrow \quad \lim _{c^{x \rightarrow \infty}} \frac{2(a x+b)}{x}=e^2 \\
& \therefore \quad e^{2 a}=e^2 \\
& \Rightarrow \quad a=1 \text { and } b \in R
\end{aligned}
$$
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