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

Answer 1

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 $0$.

Answer 2

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 $0$.