Question

limit as x tends to infinity of ln(x^2 + 2)/sqrt(x)

Answer 1

\[\lim_{x \rightarrow \infty} \frac{ x^2+2 }{ \sqrt{x} }\]Show working.

Answer 2

\[\lim_{x\to\infty}\frac{\ln(x^2+2)}{\sqrt x}=\frac{\infty}{\infty}\]
Apply L'Hopital's rule:
\[\lim_{x\to\infty}\frac{\frac{2x}{x^2+2}}{\frac{1}{2\sqrt x}}=\lim_{x\to\infty}\frac{4x^{3/2}}{x^2+2}=\frac{\infty}{\infty}\]
Apply again:
\[\lim_{x\to\infty}\frac{6x^{1/2}}{2x}=\frac{\infty}{\infty}\]
One more time:
\[\lim_{x\to\infty}\frac{3}{2\sqrt x}\]

Answer 3

So as x-> infinity,\[\frac{ 3 }{ 2\sqrt{x} }\rightarrow0\]

Answer 4

Is that right?

Answer 5

Yes

Answer 6

Thank you!

Answer 7

yw