# Question #11ebf

Sep 2, 2017

1

#### Explanation:

So we know that on the real numbers, $\sin$ and $\cos$ are bounded. This means that

$- 1 \le \cos \left(x\right) \le 1$ and $- 1 \le \sin \left(x\right) \le 1$ $\forall$ $x \in \mathbb{R}$

${\lim}_{x \rightarrow \infty} \frac{{x}^{2} + \sin \left(x\right)}{{x}^{2} + \cos \left(x\right)}$

Now divide top and bottom by ${x}^{2}$, the limit becomes:

${\lim}_{x \rightarrow \infty} \frac{1 + \frac{\sin \left(x\right)}{{x}^{2}}}{1 + \frac{\cos \left(x\right)}{{x}^{2}}}$

As x tends to infinity $\sin \left(x\right)$ and $\cos \left(x\right)$ remain bounded however ${x}^{2} \rightarrow \infty$, hence

${\lim}_{x \rightarrow \infty} \sin \frac{x}{x} ^ 2 = 0$ and ${\lim}_{x \rightarrow \infty} \cos \frac{x}{x} ^ 2 = 0$

So we have that

${\lim}_{x \rightarrow \infty} \frac{1 + \frac{\sin \left(x\right)}{{x}^{2}}}{1 + \frac{\cos \left(x\right)}{{x}^{2}}} = \frac{1}{1} = 1$