Question

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

• A. $1$
• B. $\frac{1}{2}$
• C. $2$
• D. $-2$

Answer 1

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

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

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

As $x\to \infty \frac{1}{x},\frac{x^2}{e^x},\frac{x^4}{e^{2x}}\to 0$
Hence $L=\frac{1}{2}.$