Question

Evaluate : $limi{t}_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x$

• A. $e^4$
• B. $e^2$
• C. $e^3$
• D. $e$

Answer 1

The correct option is A

$e^4$

Explanation for the correct option:

Step-1 checking indeterminate form:

$limi{t}_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x$

$$\begin{gathered} \Rightarrow limi{t}_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x \\ \Rightarrow limi{t}_{x\to \infty }{\left[\frac{x^2\left(1+{\displaystyle \frac{5}{x}}+{\displaystyle \frac{3}{x^2}}\right)}{x^2\left(1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^2}}\right)}\right]}^x\mathbf{}\mathbf{[}\mathbf{\text{take}}\mathbf{}{\mathbf{x}}^{\mathbf{2}}\mathbf{}\mathbf{\text{common in numerator and denominator}}\mathbf{]} \\ \Rightarrow {\left(\frac{\left(1+{\displaystyle \frac{5}{\infty }}+{\displaystyle \frac{3}{{\infty }^2}}\right)}{\left(1+{\displaystyle \frac{1}{\infty }}+{\displaystyle \frac{2}{{\infty }^2}}\right)}\right)}^{\infty } \\ \Rightarrow 1^{\infty }\mathbf{}\left[\because \frac{1}{\infty }=0\right] \end{gathered}$$

Step-2 Evaluating the limit :

If ${\mathbf{limit}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathbf{f}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}{\mathbf{1}}^{\mathbf{\infty }}\mathbf{,}\mathbf{}$then ${\mathbf{limit}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\mathbf{f}{\mathbf{\left(}\mathbf{x}\mathbf{\right)}}^{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{=}{{\mathbf{e}}^{{\mathbf{limit}}_{\mathbf{x}\mathbf{\to }\mathbf{\infty }}\left(f\left(x\right)-1\right)}}^{\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}$

$limi{t}_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x=e^{limi{t}_{x\to \infty }x.\left(\frac{x^2+5x+3}{x^2+x+2}-1\right)}$

first, we solve

$$\begin{gathered} limi{t}_{x\to \infty }x.\left(\frac{x^2+5x+3}{x^2+x+2}-1\right)=limi{t}_{x\to \infty }x.\left(\frac{\left(x^2+5x+3\right)-\left(x^2+x+2\right)}{x^2+x+2}\right) \\ =limi{t}_{x\to \infty }x.\left(\frac{\cancel{x^2}+5x+3-\cancel{x^2}-x-2}{x^2+x+2}\right) \\ =limi{t}_{x\to \infty }x.\left(\frac{4x-2}{x^2+x+2}\right) \end{gathered}$$

Take $x$ common in numerator and $x^2$common in denominator

$$\begin{gathered} =limi{t}_{x\to \infty }\cancel{x}.\left(\frac{\cancel{x}\left(4-{\displaystyle \frac{2}{x}}\right)}{\cancel{x^2}\left(1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^2}}\right)}\right) \\ =limi{t}_{x\to \infty }\frac{\left(4-{\displaystyle \frac{2}{x}}\right)}{\left(1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{2}{x^2}}\right)} \\ =\frac{\left(4-{\displaystyle \frac{2}{\infty }}\right)}{\left(1+{\displaystyle \frac{1}{\infty }}+{\displaystyle \frac{2}{{\infty }^2}}\right)} \\ =\frac{4}{1}\left[\because \frac{1}{\infty }=0\right] \end{gathered}$$

$limi{t}_{x\to \infty }{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x=e^4\mathbf{}\left[\because {\mathrm{limit}}_{x\to \infty }x.\left(\frac{x^2+5x+3}{x^2+x+2}-1\right)=4\right]$

Hence, option (A) is the correct answer