Question

Assertion $\left(A\right):\underset{x\to \infty }{lim}{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x=e^4$

Reason $\left(R\right):\underset{x\to \infty }{lim}{(1+x)}^{\frac{1}{x}}=e$

• A. Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
• B. Both $A$ and $R$ are true and $R$ is not the correct explanation of $A$
• C. $A$ is true but $R$ is false
• D. $R$ is true but $A$ is false

Answer 1

The correct option is C $A$ is true but $R$ is false

$\underset{x\to \infty }{\text{lim}}{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x=\underset{x\to \infty }{lim}{(\frac{x^2+5x+3}{x^2+x+2}+1-1)}^x=\underset{x\to \infty }{lim}{(1+\frac{4x+1}{x^2+x+2})}^x=\underset{x\to \infty }{lim}{(1+\frac{4x+1}{x^2+x+2})}^x=e^{\underset{x\to \infty }{lim}\frac{4x+1}{x^2+x+2}\times x}=e^4$

Let,
$L=\underset{x\to \infty }{lim}{(1+x)}^{\frac{1}{x}}\Rightarrow L=e^{\underset{x\to \infty }{lim}\frac{1}{x}\log (1+x)}\Rightarrow L=e^{\underset{x\to \infty }{lim}\frac{1}{x}\log (1+x)}\frac{\infty }{\infty }\text{form so, using L'Hospital Rule}\Rightarrow L=e^{\underset{x\to \infty }{lim}\frac{1}{1+x}}=e^0=1$