Question

For a certain value of $C$, the $\underset{x\to +\infty }{lim}(x^4+5x^3+3{)}^c-x$ is finite and non-zero quantity $L$. Find $(L-\frac{1}{4})$

Answer 1

$L=\underset{x\to \infty }{lim}{(x^4+5x^3+3)}^c-x$
$=\underset{x\to \infty }{lim}{x}^{4c}[{(1+\frac{5}{x}+\frac{3}{x^4})}^c-\frac{1}{x^{4c-1}}]$
$=\underset{x\to \infty }{lim}\frac{[{(1+\frac{5}{x}+\frac{3}{x^4})}^c-\frac{1}{x^{4c-1}}]}{\frac{1}{x^{4c}}}$
Let us consider $c=\frac{1}{4}$ because it can be any value except 0
$4c-1=0$
$c=\frac{1}{4}$
To make it simple
$\underset{x\to \infty }{lim}\frac{[{(1+\frac{5}{x}+\frac{3}{x^4})}^{\frac{1}{4}}-1]}{1/x}$
Using 'L' hospital Rule
$L=\underset{x\to \infty }{lim}\frac{\left(\frac{1}{4}\right){(1+\frac{5}{x}+\frac{3}{x^4})}^{\frac{-3}{4}}(\frac{-5}{x^2}-\frac{12}{x^5})}{\left(\frac{-1}{x^2}\right)}$

$=\underset{x\to \infty }{lim}\left(\frac{1}{4}\right){(1+\frac{5}{x}+\frac{3}{x^4})}^{\frac{-3}{4}}(5+\frac{12}{x^3})$
$\frac{1}{4}\left(1\right)\left(5\right)=L$
$L-\frac{1}{4}=1$