Question

$f\left(x\right)=\frac{\ln (x^2+e^x)}{\ln (x^4+e^{2x})}$. Then $\underset{x\to \infty }{lim}f\left(x\right)$ is equal to

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. None of these

Answer 1

The correct option is B $\frac{1}{2}$

$L=\underset{x\to \infty }{lim}\frac{\ln (x^2+e^x)}{\ln (x^4+e^{2x})}=\underset{x\to \infty }{lim}\frac{\ln e^x(1+\frac{x^2}{e^x})}{\ln e^{2x}(1+\frac{x^4}{e^{2x}})}$

$=\underset{x\to \infty }{lim}\frac{x+ln(1+\frac{x^2}{e^x})}{2x+\ln (1+\frac{x^4}{e^{2x}})}$

$=\underset{x\to \infty }{lim}\frac{1+\frac{1}{x}\ln (1+\frac{x^2}{e^x})}{2+\frac{1}{x}\ln (1+\frac{x^4}{e^{2x}})}$

Now as $x\to \infty \text{,}\frac{x^2}{e^x}\to 0\text{and as}x\to \infty ,\frac{x^4}{e^{2x}}\to 0\text{[Using L'Hospital's Rule]}$

Hence $L=\frac{1}{2}$