Calculate Limits with Trig Functions

Previously, I calculated limits (numbers that represent what the limits of certain expressions are) using the difference of squares method.

In this post, I’ll calculate a limit that contains trigonometric functions. This is useful as it combines trigonometry with limits. Specifically, I’ll explain how to calculate

$$ \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \, = \, 0 $$

The Denominator

We first focus on the limit

$$ \lim_{x \to \infty} \frac{1}{\sqrt{x}} $$

Note that as \( x \) goes to infinity, \( \sqrt{x} \) does too. Now consider \( 1 / \sqrt{x} \). The larger \( \sqrt{x} \) is, the smaller \( 1 / \sqrt{x} \) becomes.

What all this means is that as \( x \) goes to infinity, \( 1 / \sqrt{x} \) goes to zero. That is,

$$ \lim_{x \to \infty} \frac{1}{\sqrt{x}} \, = \, 0 $$

The Numerator

Next, we focus on

$$ \sin^2(x) + \cos^3(x) $$

Let’s break it down. First, we’ll look at \( \sin(x) \).

An useful property of sine is that this is always true:

$$ -1 \leq \sin(x) \leq 1 $$

The square of any number between \( -1 \) and \( 1 \) is a number between \( 0 \) and \( 1 \). For instance, \( 0.5^2 = 0.25 \) and \( (-0.1)^2 = 0.001 \).

So,

$$ 0 \leq \sin^2(x) \leq 1 $$

Next, we’ll look at \( \cos(x) \).

A useful property of cosine is that this is always true:

$$ -1 \leq \cos(x) \leq 1 $$

The cube of any number between \( -1 \) and \( 1 \) is a number between \( -1 \) and \( 1 \). For instance, \( 0.5^3 = 0.125 \) and \( (-0.1)^3 = -0.0001 \). So,

$$ -1 \leq \cos^3(x) \leq 1 $$

Since \( \sin^2(x) \geq 0 \) and \( \cos^3(x) \geq -1 \),

$$ \sin^2(x) + \cos^3(x) \geq 0 + (-1) = -1 $$

Since \( \sin^2(x) \leq 1 \) and \( \cos^3(x) \leq 1 \),

$$ \sin^2(x) + \cos^3(x) \leq 1 + 1 = 2 $$

So,

$$ -1 \leq \sin^2(x) + \cos^3(x) \leq 2 $$

Using the Numerator and the Denominator

Consider

$$ -1 \leq \sin^2(x) + \cos^3(x) \leq 2 $$

Dividing each of the above numbers by \( \sqrt{x} \), we get

$$\frac{-1}{\sqrt{x}} \leq \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \leq \frac{2}{\sqrt{x}} $$

(this works because \( \sqrt{x} \) is positive).

Recall that

$$ \lim_{x \to \infty} \frac{1}{\sqrt{x}} \, = \, 0. $$

So

$$ \lim_{x \to \infty} \frac{-1}{\sqrt{x}} \, = -1 \cdot \lim_{x \to \infty} \frac{1}{\sqrt{x}} \, = \, 0 $$

and

$$ \lim_{x \to \infty} \frac{2}{\sqrt{x}} \, = 2 \cdot \lim_{x \to \infty} \frac{1}{\sqrt{x}} \, = \, 0.$$

Since

$$\frac{-1}{\sqrt{x}} \leq \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}},$$

$$ \lim_{x \to \infty} \frac{-1}{\sqrt{x}} \leq \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}}. $$

So,

$$ 0 \leq \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}}. $$

And since

$$ \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \leq \frac{2}{\sqrt{x}}, $$

$$ \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \leq \lim_{x \to \infty} \frac{2}{\sqrt{x}}. $$

So,

$$ \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \leq 0. $$

So,

$$ 0 \leq \lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} \leq 0 $$

and the answer is

$$\lim_{x \to \infty} \frac{\sin^2(x) + \cos^3(x)}{\sqrt{x}} = 0 \,.$$