# $\lim_{x\to ∞} \sqrt{x+\sin(x)/{(x-\cos^2(x))}}$

on solving $\lim_{x\to ∞} \sqrt{x+\sin(x)/{(x-\cos^2(x))}}$. I divided by x in both numerator and denominator . and since $\lim_{x\to ∞} sin(x)/x=0$ and $\lim_{x\to ∞} cos(x)/x=0$. i arrive at $\lim_{x\to ∞} 1/\sqrt{1-0.∞}$ but this limit can not be written as equal to 1,because this is indeterminate form. so how should i proceed further?

• You actually get $\lim_{x\to\infty}\sqrt{\dfrac{1+\frac{\sin x}{x}}{1-\dfrac{\cos^2 x}{x}}}$. As you noticed, $\lim_{x\to\infty}\dfrac{\sin x}{x}=0$ and $\lim_{x\to\infty}\dfrac{\cos^2 x}{x}=0$. So, you end up with the expression $\sqrt{\dfrac{1+0}{1-0}}=\sqrt{1}=1.$ Oct 18, 2016 at 15:17
• $\lim_{x\to\infty}\dfrac{\cos x}{x}=0$ ?? no? implying that $\lim_{x\to\infty}\dfrac{\cos^2 x}{x^2}=0$ Oct 18, 2016 at 15:19
• It is true that $\lim_{x\to\infty}\dfrac{cos x}{x}=0$, but the reason (explained in the answers below) is that $\cos x$ is bounded between $-1$ and $1$ while $x$ goes to infinity. Notice that $\cos^2 x$ is also bounded (between 0 and 1) while $x$ goes to infinity, so you also have $\lim_{x\to\infty}\dfrac{\cos^2 x}{x}=0$. Oct 18, 2016 at 15:22

We have

$\forall x>0$

$\frac{|sin(x)|}{x}\leq \frac{1}{x}$

thus

$lim_{x\to +\infty}\frac{sin(x)}{x}=0$

and so for $cos$.

your limit is $\sqrt{ \frac{1+0}{1-0} }=1$.

$$\dfrac{x+\sin x}{x-\cos^2x}=\dfrac{1+\dfrac{\sin x}x}{1+\dfrac{-\cos^2x}x}$$

Now $-1\le-\cos^2x\le0,-1\le\sin x\le1$

One may observe that, as $x \to \infty$, $$\sqrt{\frac{x+\sin(x)}{x-\cos^2(x)}}=\sqrt{\frac{1+\frac{\sin(x)}x}{1-\frac{\cos^2(x)}x}} \to \sqrt{1}=1.$$

A Taylor series expansion around $x = \infty$ of your function gives $$f(x) = 1 + \frac{\cos^2 x + \sin x}{2x} + O \left( \frac{1}{x^2} \right) \, ,$$ so that the limit is 1.

With equivalents:

As $\sin x$ and $\cos x$ are bounded, $x-\sin x\sim_\infty x$, $\;x-\cos^2x\sim_\infty x$, so $$\frac{x-\sin x}{x-\cos^2x}\sim_\infty \frac xx=1.$$