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The value of \( \underset{x\xrightarrow{} \infty }{lim}{\left(\dfrac{{x}^{2}+5x+3}{{x}^{2}+x+2}\right)}^{x}\) equals
( )
A. \( {e}^{4}\)
B. \( {e}^{2}\)
C. \( {e}^{3}\)
D. \( e\)
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The value of \( \underset{x\xrightarrow{} \infty }{lim}{\left(\dfrac{{x}^{2}+5x+3}{{x}^{2}+x+2}\right)}^{x}\) equals
( )
A. \( {e}^{4}\)
B. \( {e}^{2}\)
C. \( {e}^{3}\)
D. \( e\)

Answer

A
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Solution

\(\begin{array}{l}\underset{x\to \infty }{\lim }{\left. \left(\dfrac{{x}^{2}+5x+3}{{x}^{2}+x+2}\right)\right. }^{x}={e}^{m}\\ m=\underset{x\to \infty }{\lim }\left. \left(\dfrac{{x}^{2}+5x+3}{{x}^{2}+x+2}−1\right)\right. \times x\\ m=\underset{x\to \infty }{\lim }\left. \left(\dfrac{4x+1}{{x}^{2}+x+2}\right)\right. \times x\\ =\underset{x\to \infty }{\lim }\left. \left(\dfrac{4{x}^{2}+x}{{x}^{2}+x+2}\right)\right. \\ =\underset{x\to \infty }{\lim }\left. \left(\dfrac{4+\dfrac{1}{x}}{1+\dfrac{1}{x}+\dfrac{2}{{x}^{2}}}\right)\right. \\ =\underset{x\to \infty }{\lim }\left. \left(\dfrac{4+0}{1+0+0}\right)\right. \\ =\underset{x\to \infty }{\lim }\left. \left(\dfrac{4}{1}\right)\right. \\ =\underset{x\to \infty }{\lim }\left. \left(4\right)\right. \\ =4\\ y={e}^{4}\end{array}\)
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