Question

For some values of ${}^{\prime }{k}^{\prime }$
$B=\underset{x\to \infty }{lim}{x}^k\left(\ln \right(x^2+2)-2\ln x)(e^{\frac{1}{x}}-1)$ is a finite non-zero number, then the value of $(B+k)$ is

Answer 1

$B=\underset{x\to \infty }{lim}{x}^k\left(\ln \right(x^2+2)-2\ln x)(e^{\frac{1}{x}}-1)=\underset{x\to \infty }{lim}{x}^k\left(\ln \frac{(x^2+2)}{x^2}\right)(e^{\frac{1}{x}}-1)$
$=\underset{x\to \infty }{lim}{x}^k\left\{(\frac{\ln (1+\frac{2}{x^2})}{\frac{2}{x^2}}\times \frac{2}{x^2})(\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}\times \frac{1}{x})\right\}=\underset{x\to \infty }{lim}{x}^k(1\times \frac{2}{x^2}\times 1\times \frac{1}{x})=\underset{x\to \infty }{lim}\frac{2x^k}{x^3}$
For this to be a finite non-zero number $k=3\Rightarrow B=2$