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$\lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x$ is equal to
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1735 Upvotes
Verified Answer
The correct answer is:
$e^2$
Given,
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1-1+\frac{x^2+x+3}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1+\frac{2 x+1}{x^2-x+2}\right]^x \\
& =e^{\lim _{x \rightarrow \infty} \frac{2 x+1}{x^2-x+2} \cdot(x)}
\end{aligned}
$$
$$
\begin{aligned}
& {\left[\therefore \lim _{x \rightarrow \infty}[1+f(x)]^{g(x)}=e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}\right]} \\
& =e^{\lim _{x \rightarrow \infty} \frac{2 x^2+x}{x^2-x+2}}=e^2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{x \rightarrow \infty}\left[\frac{x^2+x+3}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1-1+\frac{x^2+x+3}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]^x \\
& =\lim _{x \rightarrow \infty}\left[1+\frac{2 x+1}{x^2-x+2}\right]^x \\
& =e^{\lim _{x \rightarrow \infty} \frac{2 x+1}{x^2-x+2} \cdot(x)}
\end{aligned}
$$
$$
\begin{aligned}
& {\left[\therefore \lim _{x \rightarrow \infty}[1+f(x)]^{g(x)}=e^{\lim _{x \rightarrow \infty} f(x) \cdot g(x)}\right]} \\
& =e^{\lim _{x \rightarrow \infty} \frac{2 x^2+x}{x^2-x+2}}=e^2 \\
&
\end{aligned}
$$
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