Finding the Limit of the Trigonometric Function: A Practical Guide with Examples and Tips
When you encounter an expression like (\displaystyle \lim_{x\to 0}\frac{\sin x}{x}) or (\displaystyle \lim_{x\to \frac{\pi}{2}}\frac{\cos x}{\sin x - 1}), the first instinct is often to plug in the value directly. Day to day, understanding how to handle these situations is essential for anyone studying calculus, physics, or engineering. Also, in many cases, however, this leads to an indeterminate form such as (0/0) or (\infty/\infty). This article walks you through the most reliable methods for finding the limit of the trigonometric function, explains the underlying theory, and provides step‑by‑step examples you can practice right away.
Key Concepts and Standard Limits
Before diving into problem‑solving techniques, it’s helpful to memorize a few fundamental limits that appear repeatedly in calculus.
- (\displaystyle \lim_{x\to 0}\frac{\sin x}{x}=1)
- (\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^{2}}=\frac{1}{2})
- (\displaystyle \lim_{x\to 0}\frac{\tan x}{x}=1)
- (\displaystyle \lim_{x\to 0}\frac{\sin ax}{\sin bx}= \frac{a}{b}) (for constants (a,b\neq0))
These results are derived using geometric arguments or series expansions and serve as building blocks for more complex limits.
Step‑by‑Step Techniques
1. Direct Substitution (When It Works)
If the function is continuous at the point of interest, simply substitute the value.
Example: (\displaystyle \lim_{x\to \frac{\pi}{4}}\frac{\sin x + \cos x}{1+\tan x})
Since (\sin, \cos, \tan) are continuous at (\pi/4), we get (\frac{\frac{\sqrt2}{2}+\frac{\sqrt2}{2}}{1+1}= \frac{\sqrt2}{2}).
2. Algebraic Manipulation
When direct substitution yields (0/0) or (\infty/\infty), rewrite the expression using identities.
- Use Pythagorean identities: (\sin^{2}x + \cos^{2}x = 1).
- Factor common terms: (\frac{\sin x - \sin y}{\cos x + \cos y}) can be transformed with sum‑to‑product formulas.
Example: (\displaystyle \lim_{x\to 0}\frac{1-\cos 2x}{x^{2}})
Apply the identity (1-\cos 2x = 2\sin^{2}x):
[ \lim_{x\to 0}\frac{2\sin^{2}x}{x^{2}} = 2\left(\lim_{x\to 0}\frac{\sin x}{x}\right)^{2}=2\cdot1^{2}=2. ]
3. L’Hôpital’s Rule (For Indeterminate Forms)
If you have (\displaystyle \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{0}{0}) or (\frac{\infty}{\infty}), differentiate the numerator and denominator separately.
Example: (\displaystyle \lim_{x\to 0}\frac{\sin 3x}{x})
Both numerator and denominator tend to 0. Apply L’Hôpital:
[ \lim_{x\to 0}\frac{3\cos 3x}{1}=3\cos 0 = 3. ]
Note: L’Hôpital’s rule can be applied repeatedly if the new limit is still indeterminate.
4. Series Expansion (Taylor/Maclaurin)
Replace trigonometric functions with their polynomial approximations near the limit point.
- (\sin x = x - \frac{x^{3}}{6} + O(x^{5}))
- (\cos x = 1 - \frac{x^{2}}{2} + O(x^{4}))
- (\tan x = x + \frac{x^{3}}{3} + O(x^{5}))
Example: (\displaystyle \lim_{x\to 0}\frac{\tan x - x}{x^{3}})
Insert series:
[ \frac{(x + \frac{x^{3}}{3} + \dots) - x}{x^{3}} = \frac{\frac{x^{3}}{3} + O(x^{5})}{x^{3}} \to \frac{1}{3}. ]
5. Squeeze Theorem (Sandwich Theorem)
When a trigonometric expression is bounded between two functions whose limits are equal, the target limit equals that common value.
Example: (\displaystyle \lim_{x\to 0}\frac{x\sin x}{1})
Since (-|x| \le x\sin x \le |x|) and (\lim_{x\to 0}|x| = 0), the squeeze theorem gives 0.
Practical Examples
Below are three detailed walkthroughs that combine the techniques above.
Example 1: A Classic (\frac{0}{0}) Limit
Problem: (\displaystyle \lim_{x\to 0}\frac{\sin 5x}{\sin 2x})
Solution:
Both numerator and denominator approach 0. Use the standard limit (\frac{\sin ax}{\sin bx} \to \frac{a}{b}):
[ \lim_{x\to 0}\frac{\sin 5x}{\sin 2x}= \frac{5}{2}. ]
Alternative: Apply L’Hôpital’s rule:
[ \lim_{x\to 0}\frac{5\cos 5x}{2\cos 2x}= \frac{5}{2}. ]
Example 2: Higher‑Order Indeterminate Form
Problem: (\displaystyle \lim_{x\to 0}\frac{1-\cos x}{x^{2}})
Solution:
Use the identity (1-\cos x = 2\sin^{2}\frac{x}{2}):
[ \lim_{x\to 0}\frac{2\sin^{2}\frac{x}{2}}{x^{2}} = \frac{2}{4}\left(\lim_{x\to 0}\frac{\sin\frac{x}{2}}{\frac{x}{2}}\right)^{2}= \frac{1}{2}. ]
Example 3: Combining Multiple Techniques
Problem: (\displaystyle \lim_{x\to \frac{\pi}{2}^{-}}\frac{\sin x}{1-\cos x})
Solution:
As (x\to \frac{\pi}{2}^{-}), (\sin x\to 1) and (\cos x\to 0^{+}). The denominator tends to 1, so the limit is 1.
If you prefer an algebraic route, multiply numerator and denominator by (1+\cos x):
[ \frac{\sin x(1+\cos x)}{(1-\cos x)(1+\cos x)}
