Special limits are famous limits that appear so often in calculus that they are treated as standard tools.
The most important ones happen near \(x=0\), especially for trigonometric, exponential, and logarithmic functions.
1. The fundamental sine limit
The most important special trigonometric limit is:
\[
\lim_{x\to0}\frac{\sin x}{x}=1
\]
This limit assumes that \(x\) is measured in radians. It is not true in this simple form if degrees are used.
2. Geometric unit-circle justification
For \(0\lt x\lt\pi/2\), the unit-circle geometry gives:
\[
\sin x\le x\le\tan x
\]
Dividing by \(\sin x\) and simplifying gives:
\[
1\le\frac{x}{\sin x}\le\frac{1}{\cos x}
\]
Taking reciprocals gives:
\[
\cos x\le\frac{\sin x}{x}\le1
\]
Since \(\cos x\to1\) as \(x\to0\), the Squeeze Theorem gives:
\[
\lim_{x\to0}\frac{\sin x}{x}=1
\]
3. Tangent special limit
Since \(\tan x=\sin x/\cos x\), we have:
\[
\frac{\tan x}{x}
=
\frac{\sin x}{x}\cdot\frac{1}{\cos x}
\]
Therefore:
\[
\lim_{x\to0}\frac{\tan x}{x}
=
1\cdot1=1
\]
4. Cosine special limit
Another important special limit is:
\[
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12
\]
Use the identity:
\[
1-\cos x=2\sin^2\left(\frac{x}{2}\right)
\]
Then:
\[
\frac{1-\cos x}{x^2}
=
\frac{2\sin^2(x/2)}{x^2}
\]
Rewrite it in terms of the sine limit:
\[
\frac{2\sin^2(x/2)}{x^2}
=
\frac12
\left(\frac{\sin(x/2)}{x/2}\right)^2
\]
Since \(\sin u/u\to1\), we get:
\[
\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12
\]
5. Scaled sine limits
If \(k\) is a constant, then:
\[
\lim_{x\to0}\frac{\sin(kx)}{x}=k
\]
because:
\[
\frac{\sin(kx)}{x}
=
k\cdot\frac{\sin(kx)}{kx}
\]
and \(kx\to0\) as \(x\to0\).
6. Scaled cosine limits
Similarly:
\[
\lim_{x\to0}\frac{1-\cos(kx)}{x^2}
=
\frac{k^2}{2}
\]
This follows from substituting \(u=kx\) into the standard cosine limit.
7. Exponential special limit
A basic exponential special limit is:
\[
\lim_{x\to0}\frac{e^x-1}{x}=1
\]
More generally:
\[
\lim_{x\to0}\frac{e^{kx}-1}{x}=k
\]
For a base \(a\gt0\), \(a\ne1\):
\[
\lim_{x\to0}\frac{a^x-1}{x}=\ln a
\]
8. Logarithmic special limit
The logarithmic companion is:
\[
\lim_{x\to0}\frac{\ln(1+x)}{x}=1
\]
More generally:
\[
\lim_{x\to0}\frac{\ln(1+kx)}{x}=k
\]
This limit requires the logarithm input to stay positive near \(0\).
9. Table of common special limits
| Limit |
Value |
Main idea |
| \(\displaystyle \lim_{x\to0}\frac{\sin x}{x}\) |
\(1\) |
Unit-circle squeeze theorem |
| \(\displaystyle \lim_{x\to0}\frac{\tan x}{x}\) |
\(1\) |
\(\tan x=\sin x/\cos x\) |
| \(\displaystyle \lim_{x\to0}\frac{1-\cos x}{x^2}\) |
\(\frac12\) |
\(1-\cos x=2\sin^2(x/2)\) |
| \(\displaystyle \lim_{x\to0}\frac{e^x-1}{x}\) |
\(1\) |
Exponential standard limit |
| \(\displaystyle \lim_{x\to0}\frac{\ln(1+x)}{x}\) |
\(1\) |
Logarithmic standard limit |
| \(\displaystyle \lim_{x\to0}\frac{a^x-1}{x}\) |
\(\ln a\) |
Rewrite \(a^x=e^{x\ln a}\) |
10. Worked example
Evaluate:
\[
\lim_{x\to0}\frac{1-\cos x}{x^2}
\]
Use:
\[
1-\cos x=2\sin^2\left(\frac{x}{2}\right)
\]
Then:
\[
\frac{1-\cos x}{x^2}
=
\frac{2\sin^2(x/2)}{x^2}
=
\frac12
\left(\frac{\sin(x/2)}{x/2}\right)^2
\]
Since:
\[
\lim_{x\to0}\frac{\sin(x/2)}{x/2}=1
\]
the final result is:
\[
\lim_{x\to0}\frac{1-\cos x}{x^2}
=
\frac12
\]
11. Common mistakes
- Use radians, not degrees, for trigonometric special limits.
- Do not substitute \(x=0\) directly into \(\sin x/x\); it gives \(0/0\), but the limit exists.
- Remember that \((1-\cos x)/x\to0\), while \((1-\cos x)/x^2\to1/2\).
- For \(\sin(kx)/x\), the answer is \(k\), not always \(1\).
- For \((1-\cos(kx))/x^2\), the answer is \(k^2/2\).
- For logarithmic limits, make sure the logarithm input is positive near the approach point.
