# Calculus 2: Find limit as x approaches infinity for trig function [closed]

I am confused by the following limit:

When I graph it on my calculator I can see the limit goes towards 0, but how do I mathematically do this problem?

• $-1\leq \sin(5x) \leq 1$. Commented Feb 23, 2020 at 17:22
• Would be more interesting to take the limit of this ratio as $x\rightarrow 0.$
– mjw
Commented Feb 23, 2020 at 18:09

This is my first time posting here. Please excuse the formatting!

Since this question comes from a calculus course, you might want to learn this a little more intuitively. We know that $$-1\leq \sin(5x) \leq 1$$ (since changing the period doesn't affect the amplitude). Since we're considering $$x\to\infty$$ ("eventually" positive), we don't have to worry about flipping around the inequalities when we multiply everything by $$1/x$$. The goal now is to get the inner term into the limit form you posted:

$$-1\leq \sin(5x) \leq 1$$ $$\frac{-1}{x}\leq \frac{\sin(5x)}{x} \leq\frac{1}{x}$$ $$\lim_{x\to\infty}\frac{-1}{x}\leq\lim_{x\to\infty}\frac{\sin(5x)}{x}\leq\lim_{x\to\infty}\frac{1}{x}$$ $$0\leq\lim_{x\to\infty}\frac{\sin(5x)}{x}\leq0$$

Looking at the last inequality, the only possibility that remains is that $$\lim_{x\to\infty}\frac{\sin(5x)}{x}=0.$$

A less rigorous but more intuitive explanation is that $$\sin(5x)$$ is bounded by $$-1$$ and $$1$$, so it won't matter for extremely large values of $$x$$. The only thing that changes significantly is the denominator, so you're comparing some value oscillating between $$-1$$ and $$1$$ to some extremely large number. Therefore, the limit is $$0$$.

Note that,

$$\lim_{x\to\infty} \frac{\sin 5x}x \le \lim_{x\to\infty} \frac{1}x=0$$

Similarly,

$$\lim_{x\to\infty} \frac{\sin 5x}x \ge -\lim_{x\to\infty} \frac{1}x=0$$

Thus,

$$\lim_{x\to\infty} \frac{\sin 5x}x=0$$