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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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## QuestionMathsClass 11

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. $$e$$

A
4.6
4.6

## Solution

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