Infinite limit of trigonometric function

Question

I'm trying to find the limit of a trigonometric function as x approaches $\infty$ so I can't use the fact that : $$\lim_{x\to \infty} \frac{1}{x} = 0$$

For example this limit : $$\lim_{x\to \infty} \frac{\cos(x) - 1}{x}$$

Or

$$\lim_{x\to \infty} \frac{\sin(x)}{\lfloor x \rfloor}$$

Questions are from Thomas Calculus

I'm unsure of what you're asking. Are you attempting to prove these facts? — zz20s, Mar 04 '16 at 14:22
Possible duplicate of How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$? if you intended to have limit as $x\to 0$. — JMoravitz, Mar 04 '16 at 14:24
In the case of $x\to\infty$ the numerator is bounded while the denominator is unbounded. — JMoravitz, Mar 04 '16 at 14:26

score 3 · Answer 1 · edited Mar 04 '16 at 15:07

3

Hint: $$-1\leq \sin(x),\cos(x)\leq 1$$ so numerator is just a number oscillating from $-1$ to $1$. Some number between $-1$ and $1$ divided by $\infty$ is $0$.

edited Mar 04 '16 at 15:07

ForgotALot

3,931

answered Mar 04 '16 at 14:25

Archis Welankar

15,910

score 0 · Answer 2 · answered Mar 04 '16 at 15:24

Notice:

$$\lim_{x\to\infty}\frac{\cos(x)-1}{x}$$

Since $\cos(x)-1$ is bounded and $\lim_{x\to\infty}\frac{1}{x}=0$ this limit equals $0$.

$$\lim_{x\to\infty}\frac{\sin(x)}{\lfloor x\rfloor}$$

Since $\sin(x)$ is bounded and $\lim_{x\to\infty}\frac{1}{\lfloor x\rfloor}=0$ this limit equals $0$.

Infinite limit of trigonometric function

2 Answers2