Theoretical Recall
Taylor expansion of a function $f$ at $x_0$:
\[f(x) = f(x_0) + \frac{f'(x_0)}{1!}(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2 + \dots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + o((x-x_0)^n).\]If $x_0=0$ we obtain the Maclaurin expansion.
Rules for $o$-notation:
- $o(x^m)+o(x^m) = o(x^m)$
- $o(x^m)\cdot o(x^n) = o(x^{m+n})$
- $o(x^n)+o(x^m) = o(x^{\min{m,n}})$
- $x^n \cdot o(x^m) = o(x^{n+m})$
- $o(x^n + o(x^n)) = o(x^n)$
Maclaurin series (up to relevant order):
- $(1+x)^\alpha = 1+\alpha x+\tfrac{\alpha(\alpha-1)}{2!}x^2 + \dots + o(x^n)$
- $e^x = 1+x+\tfrac{x^2}{2!}+\tfrac{x^3}{3!}+o(x^3)$
- $\log(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}-\tfrac{x^4}{4}+o(x^4)$
- $\sin x = x-\tfrac{x^3}{3!}+\tfrac{x^5}{5!}+o(x^5)$
- $\cos x = 1-\tfrac{x^2}{2}+\tfrac{x^4}{4!}+o(x^4)$
- $\tan x = x+\tfrac{x^3}{3}+\tfrac{2}{15}x^5+o(x^5)$
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.
Exercises
Exercise 1
\(\lim_{x\to0} \frac{e^x-1-x}{x^2}\)
Solution:
$e^x = 1+x+\tfrac{x^2}{2}+o(x^2)$.
Numerator $=\tfrac{x^2}{2}+o(x^2)$.
Final Result: \(\frac{1}{2}\)
Exercise 2
\(\lim_{x\to0} \frac{\log(1+x)-x}{x^2}\)
Solution:
$\log(1+x)=x-\tfrac{x^2}{2}+o(x^2)$.
Numerator $=-\tfrac{x^2}{2}+o(x^2)$.
Final Result: \(-\frac{1}{2}\)
Exercise 3
\(\lim_{x\to0} \frac{\sin x - x}{x^3}\)
Solution:
$\sin x = x-\tfrac{x^3}{6}+o(x^3)$.
Numerator $=-\tfrac{x^3}{6}+o(x^3)$.
Final Result: \(-\frac{1}{6}\)
Exercise 4
\(\lim_{x\to0} \frac{1-\cos x}{x^2}\)
Solution:
$\cos x=1-\tfrac{x^2}{2}+o(x^2)$.
Numerator $=\tfrac{x^2}{2}+o(x^2)$.
Final Result: \(\frac{1}{2}\)
Exercise 5
\(\lim_{x\to0} \frac{e^{2x}-1-2x}{x^2}\)
Solution:
$e^{2x}=1+2x+2x^2+o(x^2)$.
Numerator $=2x^2+o(x^2)$.
Final Result: \(2\)
Exercise 6
\(\lim_{x\to0} \frac{\tan x - x}{x^3}\)
Solution:
$\tan x = x+\tfrac{x^3}{3}+o(x^3)$.
Numerator $=\tfrac{x^3}{3}+o(x^3)$.
Final Result: \(\frac{1}{3}\)
Exercise 7
\(\lim_{x\to0} \frac{\log(1+x)-\sin x}{x^3}\)
Solution:
$\log(1+x)=x-\tfrac{x^2}{2}+\tfrac{x^3}{3}+o(x^3)$,
$\sin x = x-\tfrac{x^3}{6}+o(x^3)$.
Difference $=-\tfrac{x^2}{2}+\tfrac{x^3}{2}+o(x^3)$.
Dividing by $x^3$: $\sim -\tfrac{1}{2}\cdot \tfrac{1}{x}+\tfrac{1}{2}+o(1)$.
Thus, as $x\to0^+$, limit $\to -\infty$; as $x\to0^-$, limit $\to +\infty$.
Final Result:
The two-sided limit does not exist.
Exercise 8
\(\lim_{x\to0} \frac{e^x-\cos x}{x}\)
Solution:
$e^x=1+x+\tfrac{x^2}{2}+o(x^2)$,
$\cos x=1-\tfrac{x^2}{2}+o(x^2)$.
Numerator $=x+x^2+o(x^2)$. Divide by $x$: $1+x+o(x)$.
Final Result: \(1\)
Exercise 9
\(\lim_{x\to0} \frac{\sin x - \tan x}{x^3}\)
Solution:
$\sin x=x-\tfrac{x^3}{6}+o(x^3)$,
$\tan x=x+\tfrac{x^3}{3}+o(x^3)$.
Difference $=-\tfrac{x^3}{2}+o(x^3)$.
Final Result: \(-\frac{1}{2}\)
Exercise 10
\(\lim_{x\to0} \frac{e^x-\sin x-1}{x}\)
Solution:
$e^x=1+x+\tfrac{x^2}{2}+o(x^2)$,
$\sin x=x-\tfrac{x^3}{6}+o(x^3)$.
Numerator $=\tfrac{x^2}{2}+\tfrac{x^3}{6}+o(x^2)$.
Divided by $x$: $\tfrac{x}{2}+o(x)\to0$.
Final Result: \(0\)