Limits Involving Trigonometric Functions (Foundation)
In this reading, we begin building a toolkit for trigonometric limits in calculus. Up to this point, most of our limits have involved polynomials, rational functions, and simple algebraic expressions. Once we start differentiating and integrating trigonometric functions, two special limits become central:
- ###\lim_{x \to 0} \frac{\sin x}{x} = 1###
- ###\lim_{x \to 0} \frac{\tan x}{x} = 1###
These two results are the foundation of almost all basic trigonometric limits in calculus. In this reading we will:
- Understand what these limits mean and why they matter.
- Use them to evaluate more complicated trigonometric limits in calculus.
- Highlight a crucial point: these standard limits hold when the angle is measured in radians, not degrees.
1. Why do we need special trigonometric limits?
For algebraic functions, rules of limits often allow direct substitution or simple algebraic manipulation. With trigonometric limits in calculus, especially near zero, direct substitution frequently gives indeterminate forms like ##\frac{0}{0}##. For example:
###\lim_{x \to 0} \frac{\sin x}{x}.###
Substituting ##x = 0## gives
###\frac{\sin 0}{0} = \frac{0}{0},###
an indeterminate form. The limit still exists, but we must reason more carefully. Once we accept or prove that this limit is 1 (using geometric or analytic arguments), a wide family of other trigonometric limits in calculus becomes manageable.
2. The key limit: \(\displaystyle \lim_{x \to 0} \frac{\sin x}{x} = 1\)
The first and most important result is
###\lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{(with x in radians)}.###
Consider a unit circle (radius 1). For a small positive angle ##x## in radians, look at:
- The length of the arc on the circle: ##x## (since arc length = radius × angle and radius = 1).
- The vertical coordinate of the point on the circle: ##\sin x##.
For small angles, the arc length and the vertical height are very close. Geometry shows that for small positive ##x##:
###\sin x \le x \le \tan x.###
Dividing each term by ##\sin x## (for ##x \ne 0##) gives
###1 \le \frac{x}{\sin x} \le \frac{\tan x}{\sin x} = \frac{\sin x / \cos x}{\sin x} = \frac{1}{\cos x}.###
Taking reciprocals reverses the inequalities:
###\cos x \le \frac{\sin x}{x} \le 1.###
As ##x \to 0##, we know that ##\cos x \to 1##. By the squeeze theorem,
###\lim_{x \to 0} \frac{\sin x}{x} = 1.###
This result is the geometric heart of trigonometric limits in calculus.
3. From \(\sin x / x\) to \(\tan x / x\)
Once we know that
###\lim_{x \to 0} \frac{\sin x}{x} = 1,###
we can quickly deduce another key result:
###\lim_{x \to 0} \frac{\tan x}{x} = 1.###
We write
###\frac{\tan x}{x} = \frac{\sin x}{x \cos x} = \left(\frac{\sin x}{x}\right)\left(\frac{1}{\cos x}\right).###
As ##x \to 0##,
- ###\frac{\sin x}{x} \to 1,###
- ###\cos x \to 1###, so ###\frac{1}{\cos x} \to 1.###
Therefore,
###\lim_{x \to 0} \frac{\tan x}{x} = 1 \times 1 = 1.###
4. Useful derived trigonometric limits
From these two fundamental facts, we can derive several very useful limits.
4.1 \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{ax}\)
Let ##a## be a non-zero constant. Consider
###\lim_{x \to 0} \frac{\sin (ax)}{ax}.###
Put ##t = ax##. As ##x \to 0##, we also have ##t \to 0## and
###\frac{\sin (ax)}{ax} = \frac{\sin t}{t}.###
Therefore,
###\lim_{x \to 0} \frac{\sin (ax)}{ax} = \lim_{t \to 0} \frac{\sin t}{t} = 1.###
4.2 \(\displaystyle \lim_{x \to 0} \frac{\sin ax}{bx}\)
Now consider
###\lim_{x \to 0} \frac{\sin (ax)}{bx}.###
We write
###\frac{\sin (ax)}{bx} = \frac{a}{b} \cdot \frac{\sin (ax)}{ax}.###
From the previous result, the limit is
###\lim_{x \to 0} \frac{\sin (ax)}{bx} = \frac{a}{b} \cdot 1 = \frac{a}{b}.###
4.3 \(\displaystyle \lim_{x \to 0} \frac{1 – \cos x}{x^2}\)
This limit is also very important:
###\lim_{x \to 0} \frac{1 – \cos x}{x^2} = \frac{1}{2}.###
Using the identity
###1 – \cos x = 2 \sin^2 \left(\frac{x}{2}\right),###
we obtain
###\frac{1 – \cos x}{x^2} = \frac{2 \sin^2 (x/2)}{x^2}
= 2 \cdot \left(\frac{\sin (x/2)}{x/2}\right)^2 \cdot \frac{1}{4}
= \frac{1}{2} \left(\frac{\sin (x/2)}{x/2}\right)^2.###
As ##x \to 0##, ##x/2 \to 0## and
###\frac{\sin (x/2)}{x/2} \to 1.###
Therefore,
###\lim_{x \to 0} \frac{1 – \cos x}{x^2} = \frac{1}{2}.###
5. Worked examples on basic trig limits
We now work through a series of examples, moving from simple to more challenging problems. Each example is fully solved.
Example 1
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 3x}{x}.###
Solution.
We write
###\frac{\sin 3x}{x} = 3 \cdot \frac{\sin 3x}{3x}.###
As ##x \to 0##, ##\frac{\sin 3x}{3x} \to 1##. Hence
###\lim_{x \to 0} \frac{\sin 3x}{x} = 3.###
Example 2
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 5x}{4x}.###
Solution.
We write
###\frac{\sin 5x}{4x} = \frac{5}{4} \cdot \frac{\sin 5x}{5x}.###
Thus
###\lim_{x \to 0} \frac{\sin 5x}{4x} = \frac{5}{4} \cdot 1 = \frac{5}{4}.###
Example 3
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\tan 2x}{x}.###
Solution.
We write
###\frac{\tan 2x}{x} = 2 \cdot \frac{\tan 2x}{2x}.###
Since ###\lim_{x \to 0} \frac{\tan 2x}{2x} = 1###, the required limit is
###\lim_{x \to 0} \frac{\tan 2x}{x} = 2.###
Example 4
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\tan 7x}{3x}.###
Solution.
We have
###\frac{\tan 7x}{3x} = \frac{7}{3} \cdot \frac{\tan 7x}{7x}.###
Hence
###\lim_{x \to 0} \frac{\tan 7x}{3x} = \frac{7}{3}.###
Example 5
Evaluate ###\displaystyle \lim_{x \to 0} \frac{1 – \cos 2x}{x^2}.###
Solution.
Using the formula
###\lim_{x \to 0} \frac{1 – \cos (ax)}{x^2} = \frac{a^2}{2},###
with ##a = 2##, we obtain
###\lim_{x \to 0} \frac{1 – \cos 2x}{x^2} = \frac{2^2}{2} = 2.###
Example 6
Evaluate ###\displaystyle \lim_{x \to 0} \frac{1 – \cos 3x}{x^2}.###
Solution.
Using the same formula with ##a = 3##, we have
###\lim_{x \to 0} \frac{1 – \cos 3x}{x^2} = \frac{3^2}{2} = \frac{9}{2}.###
Example 7
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin x – x \cos x}{x^3}.###
Solution.
We rearrange:
###\sin x – x \cos x = \sin x – x(\cos x – 1 + 1)
= (\sin x – x) + x(1 – \cos x).###
So
###\frac{\sin x – x \cos x}{x^3}
= \frac{\sin x – x}{x^3} + \frac{x(1 – \cos x)}{x^3}
= \frac{\sin x – x}{x^3} + \frac{1 – \cos x}{x^2}.###
Using the known behaviour (from expansions or advanced results),
###\lim_{x \to 0} \frac{\sin x – x}{x^3} = -\frac{1}{6}, \quad
\lim_{x \to 0} \frac{1 – \cos x}{x^2} = \frac{1}{2}.###
Hence
###\lim_{x \to 0} \frac{\sin x – x \cos x}{x^3} = -\frac{1}{6} + \frac{1}{2} = \frac{1}{3}.###
Example 8
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 5x – 5x}{x^3}.###
Solution.
Near zero, the expansion of ##\sin 5x## begins as
###\sin 5x \approx 5x – \frac{(5x)^3}{6}.###
Therefore,
###\sin 5x – 5x \approx -\frac{125 x^3}{6}.###
Thus
###\frac{\sin 5x – 5x}{x^3} \approx -\frac{125}{6},###
and the exact limit is
###\lim_{x \to 0} \frac{\sin 5x – 5x}{x^3} = -\frac{125}{6}.###
Example 9
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\tan x – \sin x}{x^3}.###
Solution.
We use the identity ###\tan x = \frac{\sin x}{\cos x}###:
###\tan x – \sin x = \sin x \left(\frac{1}{\cos x} – 1\right)
= \sin x \cdot \frac{1 – \cos x}{\cos x}.###
Hence
###\frac{\tan x – \sin x}{x^3}
= \frac{\sin x}{x} \cdot \frac{1 – \cos x}{x^2} \cdot \frac{1}{\cos x}.###
As ##x \to 0##,
- ###\frac{\sin x}{x} \to 1,###
- ###\frac{1 – \cos x}{x^2} \to \frac{1}{2},###
- ###\frac{1}{\cos x} \to 1.###
Therefore,
###\lim_{x \to 0} \frac{\tan x – \sin x}{x^3} = 1 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}.###
Example 10
Evaluate ###\displaystyle \lim_{x \to 0} \frac{x – \sin x \cos x}{x^3}.###
Solution.
Rewrite the numerator:
###x – \sin x \cos x = x – \sin x + \sin x – \sin x \cos x
= (x – \sin x) + \sin x(1 – \cos x).###
Then
###\frac{x – \sin x \cos x}{x^3}
= \frac{x – \sin x}{x^3} + \frac{\sin x(1 – \cos x)}{x^3}.###
We know
###\lim_{x \to 0} \frac{x – \sin x}{x^3} = \frac{1}{6}###
(negative of the earlier limit), and
###\frac{\sin x(1 – \cos x)}{x^3}
= \left(\frac{\sin x}{x}\right)\left(\frac{1 – \cos x}{x^2}\right)
\to 1 \cdot \frac{1}{2} = \frac{1}{2}.###
Hence
###\lim_{x \to 0} \frac{x – \sin x \cos x}{x^3} = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}.###
Example 11
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 4x}{\tan 3x}.###
Solution.
We write
###\frac{\sin 4x}{\tan 3x}
= \frac{\sin 4x}{4x} \cdot \frac{4x}{3x} \cdot \frac{3x}{\tan 3x}.###
As ##x \to 0##,
- ###\frac{\sin 4x}{4x} \to 1,###
- ###\frac{3x}{\tan 3x} \to 1,###
- ###\frac{4x}{3x} = \frac{4}{3}.###
Therefore,
###\lim_{x \to 0} \frac{\sin 4x}{\tan 3x} = 1 \cdot \frac{4}{3} \cdot 1 = \frac{4}{3}.###
Example 12
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 2x \cdot (1 – \cos 3x)}{x^3}.###
Solution.
We separate the terms:
###\frac{\sin 2x \cdot (1 – \cos 3x)}{x^3}
= \left(\frac{\sin 2x}{2x}\right)\left(\frac{1 – \cos 3x}{x^2}\right)\cdot \frac{2}{1}.###
Using the standard limits,
- ###\frac{\sin 2x}{2x} \to 1,###
- ###\frac{1 – \cos 3x}{x^2} \to \frac{9}{2}.###
Thus
###\lim_{x \to 0} \frac{\sin 2x \cdot (1 – \cos 3x)}{x^3}
= 1 \cdot \frac{9}{2} \cdot 2 = 9.###
6. Radians versus degrees: a crucial point
The standard limits
###\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{\tan x}{x} = 1###
are valid only when x is in radians. If angles are measured in degrees, the values change.
Example 13 (degrees)
Let ##\alpha## be an angle in degrees. Consider
###\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha}.###
We convert to radians:
###\alpha \ (\text{degrees}) = \frac{\pi}{180} \alpha \ (\text{radians}).###
Let ##x = \frac{\pi}{180}\alpha##. Then ##\alpha = \frac{180}{\pi}x## and
###\frac{\sin \alpha}{\alpha} = \frac{\sin x}{\frac{180}{\pi}x} = \frac{\pi}{180} \cdot \frac{\sin x}{x}.###
As ##\alpha \to 0##, ##x \to 0## and
###\lim_{\alpha \to 0} \frac{\sin \alpha}{\alpha} = \frac{\pi}{180} \cdot 1 = \frac{\pi}{180}.###
This shows that in degrees, the limit is not 1, but ##\frac{\pi}{180}##. This is why, in trigonometric limits in calculus, radians are the natural choice.
7. Additional worked examples
Example 14
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 5x}{x \cos 2x}.###
Solution.
We write
###\frac{\sin 5x}{x \cos 2x}
= \frac{\sin 5x}{5x} \cdot \frac{5}{\cos 2x}.###
As ##x \to 0##, ###\frac{\sin 5x}{5x} \to 1### and ###\cos 2x \to 1###. Therefore,
###\lim_{x \to 0} \frac{\sin 5x}{x \cos 2x} = 5.###
Example 15
Evaluate ###\displaystyle \lim_{x \to 0} \frac{\sin 3x – \sin x}{x^3}.###
Solution.
Using the identity ###\sin A – \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A – B}{2}###, we have
###\sin 3x – \sin x = 2 \cos 2x \sin x.###
Hence
###\frac{\sin 3x – \sin x}{x^3}
= \frac{2 \cos 2x \sin x}{x^3}
= 2 \cos 2x \cdot \frac{\sin x}{x} \cdot \frac{1}{x^2}.###
As ##x \to 0##, ###\cos 2x \to 1### and ###\frac{\sin x}{x} \to 1###, while ###\frac{1}{x^2} \to \infty###, so the limit diverges to infinity. The expression grows without bound as ##x \to 0##.
8. References for further reading
- The article on limits at Wikipedia provides a broad overview of limit concepts and examples.
- Introductory notes from organisations such as the American Mathematical Society (AMS) and the National Institute of Standards and Technology (NIST) often discuss foundational ideas related to trigonometric functions and their limits.
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