Question

Solve:
$\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-3\sqrt{x^3+1}}{4\sqrt{x^4+1}+5\sqrt{x^4+1}}$

Answer 1

To solve : $\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-3\sqrt{x^3+1}}{4\sqrt{x^4+1}+5\sqrt{x^4+1}}$

$=\underset{x\to \infty }{lim}\frac{\sqrt{x^2+1}-3\sqrt{x^3+1}}{9\sqrt{x^4+1}}$

$=\underset{x\to \infty }{lim}\frac{\sqrt{x^2(1+\frac{1}{x^2})-3\sqrt{x^3(1+\frac{1}{x^3})}}}{9\sqrt{x^4+(1+\frac{1}{x^2})}}$

$=\underset{x\to \infty }{lim}\left(\frac{\sqrt{1+\frac{1}{x}}-3x\sqrt{x}\sqrt{1+\frac{1}{x^3}}}{9x^2\sqrt{1+\frac{1}{x^4}}}\right)$

$=\underset{x\to \infty }{lim}\left(\frac{\sqrt{1+\frac{1}{x^2}}-3\sqrt{3}\sqrt{1+\frac{1}{x^2}}}{9x\sqrt{1+\frac{1}{x^4}}}\right)$

$=\underset{x\to \infty }{lim}\left(\frac{\frac{1}{x}\sqrt{1+\frac{1}{x^2}}-\frac{3}{\sqrt{x}}\sqrt{1+\frac{1}{\infty }}}{9\sqrt{1+\frac{1}{\infty }}}\right)$

$=\frac{\frac{1}{\infty }\sqrt{1+\frac{1}{\infty }}\frac{-3}{\sqrt{\infty }}\sqrt{1+\frac{1}{\infty }}}{9\sqrt{1+\frac{1}{\infty }}}$

$=\frac{0\sqrt{1+0}-0\sqrt{1+0}}{9\sqrt{1+0}}$

$=\frac{0-0}{9}=0$