0
$\begingroup$

Evaluate $$\lim_{x\to+\infty} \frac{\sqrt{x}(\sin{x}+\sqrt{x}\cos{x})}{x\sqrt{x}-\sin({x\sqrt{x})}}$$

My attempt: $$\lim_{x\to+\infty} \frac{\sqrt{x}(\sin{x}+\sqrt{x}\cos{x})}{x\sqrt{x}-\sin({x\sqrt{x})}}=\lim_{x\to+\infty} \frac{\sin{x}+\sqrt{x}\cos{x}}{x-\frac{\sin({x\sqrt{x})}}{\sqrt{x}}}$$

I see it is possible to apply the squeeze theorem to the negative term in the denominator, but I do not know about if this is the right path. Any hint?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

HINT

You should factor $x\sqrt{x}$ instead of $\sqrt{x}$. Precisely, \begin{align*} \frac{\sqrt{x}(\sin(x) + \sqrt{x}\cos(x))}{x\sqrt{x} - \sin(x\sqrt{x})} = \frac{\frac{\sin(x)}{x} + \frac{\cos(x)}{\sqrt{x}}}{1 - \frac{\sin(x\sqrt{x})}{x\sqrt{x}}} \end{align*}

Can you take it from here?

$\endgroup$
1
  • $\begingroup$ Yes, sure. Thanks! $\endgroup$
    – Mauricio Mendes
    Jul 7, 2020 at 1:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .