Limits Using Taylor Expansions: Formulas and Solved Examples
Taylor expansions turn many indeterminate limits into algebraic calculations. The essential step is to expand each function far enough to identify the first nonzero term that remains after cancellation.
This page collects the Maclaurin formulas most often used in limits, the main rules of little-o notation, and ten solved examples involving exponential, logarithmic, and trigonometric functions.
How to Evaluate Limits Using Taylor Expansions
If $f$ is sufficiently differentiable near $x_0$, its Taylor expansion at $x_0$ is:
\[f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \dots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + o((x-x_0)^n).\]If $x_0=0$, we obtain the Maclaurin expansion.
Rules for Little-o Notation
\[o(x^m)+o(x^m)=o(x^m)\] \[o(x^m)\cdot o(x^n)=o(x^{m+n})\] \[o(x^n)+o(x^m)=o(x^{\min\{m,n\}})\] \[x^n\cdot o(x^m)=o(x^{n+m})\]For any nonzero constant $C$, $C\,o(x^n)=o(x^n)$.
Taylor and Maclaurin Expansions Commonly Used in Limits
Up to the relevant order:
\[(1+x)^\alpha = 1+\alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \dots + o(x^n)\] \[e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)\] \[\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+o(x^4)\] \[\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5)\] \[\cos x = 1-\frac{x^2}{2}+\frac{x^4}{4!}+o(x^4)\] \[\tan x = x+\frac{x^3}{3}+\frac{2}{15}x^5+o(x^5)\]Author’s note:
The expansions must be truncated only after ensuring the approximation order is sufficient to determine the limit. A common mistake is cutting the series too early.
Solved Limits Using Taylor Expansions
Exercise 1
\[\lim_{x\to0} \frac{e^x-1-x}{x^2}\]Solution.
Using the Maclaurin expansion:
\[e^x = 1+x+\frac{x^2}{2}+o(x^2).\]Therefore:
\[e^x-1-x = \frac{x^2}{2}+o(x^2).\]Dividing by x²:
\[\frac{e^x-1-x}{x^2} = \frac{1}{2}+o(1).\]Final Result
\[\frac{1}{2}\]Exercise 2
\[\lim_{x\to0} \frac{\log(1+x)-x}{x^2}\]Solution.
Using:
\[\log(1+x) = x-\frac{x^2}{2}+o(x^2),\]we obtain:
\[\log(1+x)-x = -\frac{x^2}{2}+o(x^2).\]Therefore:
\[\frac{\log(1+x)-x}{x^2} = -\frac{1}{2}+o(1).\]Final Result
\[-\frac{1}{2}\]Exercise 3
\[\lim_{x\to0} \frac{\sin x-x}{x^3}\]Solution.
Using:
\[\sin x = x-\frac{x^3}{6}+o(x^3),\]we obtain:
\[\sin x-x = -\frac{x^3}{6}+o(x^3).\]Therefore:
\[\frac{\sin x-x}{x^3} = -\frac{1}{6}+o(1).\]Final Result
\[-\frac{1}{6}\]Exercise 4
\[\lim_{x\to0} \frac{1-\cos x}{x^2}\]Solution.
Using:
\[\cos x = 1-\frac{x^2}{2}+o(x^2),\]we obtain:
\[1-\cos x = \frac{x^2}{2}+o(x^2).\]Therefore:
\[\frac{1-\cos x}{x^2} = \frac{1}{2}+o(1).\]Final Result
\[\frac{1}{2}\]Exercise 5
\[\lim_{x\to0} \frac{e^{2x}-1-2x}{x^2}\]Solution.
Using the exponential expansion with argument 2x:
\[e^{2x} = 1+2x+\frac{(2x)^2}{2}+o(x^2).\]Hence:
\[e^{2x} = 1+2x+2x^2+o(x^2).\]Therefore:
\[e^{2x}-1-2x = 2x^2+o(x^2).\]Dividing by x²:
\[\frac{e^{2x}-1-2x}{x^2} = 2+o(1).\]Final Result
\[2\]Exercise 6
\[\lim_{x\to0} \frac{\tan x-x}{x^3}\]Solution.
Using:
\[\tan x = x+\frac{x^3}{3}+o(x^3),\]we obtain:
\[\tan x-x = \frac{x^3}{3}+o(x^3).\]Therefore:
\[\frac{\tan x-x}{x^3} = \frac{1}{3}+o(1).\]Final Result
\[\frac{1}{3}\]Exercise 7
\[\lim_{x\to0} \frac{\log(1+x)-\sin x}{x^3}\]Solution.
Use the expansions:
\[\log(1+x) = x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3),\]and:
\[\sin x = x-\frac{x^3}{6}+o(x^3).\]Subtracting:
\[\log(1+x)-\sin x = -\frac{x^2}{2} + \frac{x^3}{2} + o(x^3).\]Dividing by x³:
\[\frac{\log(1+x)-\sin x}{x^3} = -\frac{1}{2x} + \frac{1}{2} + o(1).\]As x → 0⁺:
\[-\frac{1}{2x} + \frac{1}{2} + o(1) \to -\infty.\]As x → 0⁻:
\[-\frac{1}{2x} + \frac{1}{2} + o(1) \to +\infty.\]The two one-sided limits are different.
Final Result
\[\text{The two-sided limit does not exist.}\]Exercise 8
\[\lim_{x\to0} \frac{e^x-\cos x}{x}\]Solution.
Use:
\[e^x = 1+x+\frac{x^2}{2}+o(x^2),\]and:
\[\cos x = 1-\frac{x^2}{2}+o(x^2).\]Subtracting:
\[e^x-\cos x = x+x^2+o(x^2).\]Dividing by x:
\[\frac{e^x-\cos x}{x} = 1+x+o(x).\]Therefore:
\[1+x+o(x)\to1.\]Final Result
\[1\]Exercise 9
\[\lim_{x\to0} \frac{\sin x-\tan x}{x^3}\]Solution.
Use:
\[\sin x = x-\frac{x^3}{6}+o(x^3),\]and:
\[\tan x = x+\frac{x^3}{3}+o(x^3).\]Subtracting:
\[\sin x-\tan x = -\frac{x^3}{6} - \frac{x^3}{3} + o(x^3).\]Hence:
\[\sin x-\tan x = -\frac{x^3}{2}+o(x^3).\]Therefore:
\[\frac{\sin x-\tan x}{x^3} = -\frac{1}{2}+o(1).\]Final Result
\[-\frac{1}{2}\]Exercise 10
\[\lim_{x\to0} \frac{e^x-\sin x-1}{x}\]Solution.
Use:
\[e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3),\]and:
\[\sin x = x-\frac{x^3}{6}+o(x^3).\]Therefore:
\[e^x-\sin x-1 = \frac{x^2}{2} + \frac{x^3}{3} + o(x^3).\]Dividing by x:
\[\frac{e^x-\sin x-1}{x} = \frac{x}{2} + \frac{x^2}{3} + o(x^2).\]As x → 0:
\[\frac{x}{2} + \frac{x^2}{3} + o(x^2) \to0.\]Final Result
\[0\]Continue Exploring Limits
Taylor expansions are particularly useful when several terms cancel and the dominant order of an expression is not immediately visible.
The key is to expand each function far enough to identify the first nonzero term that survives the cancellation.
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