How do you find lim_(x->∞)(x+2sinx-1)/(x+3cosx+1)?

How do you find lim_(x->∞)(x+2sinx-1)/(x+3cosx+1)?

Jan 18, 2018

${\lim}_{x \rightarrow + \infty} \frac{x + 2 \sin x - 1}{x + 3 \cos x + 1} = 1$

Explanation:

${\lim}_{x \rightarrow + \infty} \frac{x + 2 \sin x - 1}{x + 3 \cos x + 1}$

$x \to + \infty$
$x > 0$

$= {\lim}_{x \rightarrow + \infty} \frac{1 + 2 \sin \frac{x}{x} - \frac{1}{x}}{1 + 3 \cos \frac{x}{x} + \frac{1}{x}} = 1$

because

• ${\lim}_{x \rightarrow + \infty} \frac{1}{x} {=}^{\left(\frac{1}{+ \infty}\right)} 0$

• ${\lim}_{x \rightarrow + \infty} \sin \frac{x}{x} = 0$

$| \sin \frac{x}{x} | \le \frac{1}{|} x |$
so

$- \frac{1}{|} x | \le \sin \frac{x}{x} \le \frac{1}{|} x |$

Using the sandwich/squeeze theorem we get:

${\lim}_{x \rightarrow + \infty} - \frac{1}{|} x | = 0 = {\lim}_{x \rightarrow + \infty} \frac{1}{|} x |$

Therefore ,

${\lim}_{x \rightarrow + \infty} \sin \frac{x}{x} = 0$

• ${\lim}_{x \rightarrow + \infty} \cos \frac{x}{x} = 0$

$| \cos \frac{x}{x} | \le \frac{1}{|} x |$

$- \frac{1}{|} x | \le \cos \frac{x}{x} \le \frac{1}{|} x |$

${\lim}_{x \rightarrow + \infty} - \frac{1}{|} x | = 0 = {\lim}_{x \rightarrow + \infty} \frac{1}{|} x |$

Therefore,

${\lim}_{x \rightarrow + \infty} \cos \frac{x}{x} = 0$