If f(x)=x2+5x+3x2+x+2x , then limx→∞ f(x) is

f(x)=1+4x+1x2+x+2x2+x+24x+1×(4x+1)x2+x+1=(1+u)u14×4+1/x1+1/x+2/x2,u=4x+1x2+x+2Since u→0 as x→∞ so limx→∞ f(x)=e4

If f(x)=x2+5x+3x2+x+2x , then limx→∞ f(x) is

f(x)=1+4x+1x2+x+2x2+x+24x+1×(4x+1)x2+x+1=(1+u)u14×4+1/x1+1/x+2/x2,u=4x+1x2+x+2Since u→0 as x→∞ so limx→∞ f(x)=e4

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If $f (x) = {(\frac{x^{2} + 5 x + 3}{x^{2} + x + 2})}^{x}$ , then $lim_{x \to \infty} f (x)$ is

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$e^{4}$

$e^{3}$

$e^{2}$

$2^{4}$

answer is A.

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Detailed Solution

$\begin{aligned} f (x) = {(1 + \frac{4 x + 1}{x^{2} + x + 2})}^{\frac{x^{2} + x + 2}{4 x + 1} \times \frac{(4 x + 1)}{x^{2} + x + 1}} \\ = (1 + u)_{u}^{\frac{1}{4} \times \frac{4 + 1 / x}{1 + 1 / x + 2 / x^{2}}}, u = \frac{4 x + 1}{x^{2} + x + 2} \end{aligned}$