Taylor's Formula.
Definition. Let the function $f(z)$ be defined in a neighborhood of the point $x_0$ and have derivatives up to $(n-1)$th order inclusive in this neighborhood, and let $f^{(n)}(x_0)$ exist. Then
$$f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+...+$$ $$+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+o((x-x_0)^n)$$ при $x\rightarrow x_0. $
So $$f(x)=\sum\limits_{k=0}^{n}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k+o((x-x_0)^n),\qquad x\rightarrow x_0.$$
The polynomial $P_n(x)=\sum\limits_{k=0}^{n}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k $ is the Taylor polynomial. $r_n=f(x)-P_n(x)$ is the remainder term of the $n$th order Taylor formula.
For $x_0=0$, $f(x)=\sum\limits_{k=0}^n\frac{f^{(k)}(0)}{k!}x^k+o(x^n)$ - the Maclaurin formula.
Taylor formulas around the point $x_0=0$ for basic elementary functions.
$$e^x=1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+o(x^n);\quad x\rightarrow 0$$
$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}+...+\frac{(-1)^nx^{2n+1}}{(2n+1)!}+o(x^{2n+2}); \quad x\rightarrow 0$$
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...+(-1)^n\frac{x^{2n}}{(2n)!}+o(x^{2n+1});\quad x\rightarrow 0$$
$$(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+...+$$ $$+\frac{\alpha(\alpha-1)...(\alpha-(n-1))}{n!}x^n+o(x^n);\quad x\rightarrow 0$$
$$\frac{1}{1+x}=\sum\limits_{k=0}^n (-1)^kx^k+o(x^n);\quad x\rightarrow 0$$
$$\frac{1}{1-x}=\sum\limits_{k=0}^n x^k+o(x^n);\quad x\rightarrow 0$$
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}+...+\frac{(-1)^{n-1}x^n}{n}+o(x^n);\quad x\rightarrow 0$$
$$\ln(1-x)=-\sum\limits_{k=0}^n \frac{x^k}{k}+o(x^n)\quad x\rightarrow 0.$$
$$tg x = x+\frac{x^3}{3}+\frac{x^5}{15}+\overline{o}(x^5) \quad x\rightarrow 0$$
$$\arcsin x = x+\sum\limits_{k=1}^n\frac{(2k-1)!!}{2^k k!(2k+1)}x^{2k+1}+\overline{o}(x^{2n+2}), \quad x\rightarrow 0$$
Examples.
Expand using the Maclaurin formula to $o(x^n)$ the function
${\bf 1. e^{5x-1}}$
Solution.
$$e^{5x-1}=\frac{e^{5x}}{e}=\sum\limits_{k=0}^n\frac{(5x)^k}{k!e}+o(x^n).$$
${\bf 2. \sin(2x+3)}$
Solution.
$$\sin(2x+3)=\sin 2x\cos 3+\sin 3\cos 2x=$$
$$\cos 3\sum\limits_{k=0}^n(-1)^k\frac{(2x)^{2k+1}}{(2k+1)!}+\sin 3\sum\limits_{k=0}^n(-1)^k\frac{(2x)^{2k}}{(2k)!}+o(x^{2n+1})$$
${\bf 3.}\cos\left(\frac{x}{2}+2\right)$
Solution.
$$\cos^{(k)}\left(\frac{x}{2}+2\right)=\frac{1}{2^k}\cos\left(\frac{x}{2}+2+\frac{\pi}{2}k\right)$$
$$\cos\left(\frac{x}{2}+2\right)=\sum\limits_{k=0}^n\frac{1}{2^k}\frac{\cos(2+\frac{\pi}{2}k)}{k!}x^k+o(x^k).$$
${\bf 4. \frac{1}{1-2x}}$
Solution.
$$\frac{1}{1-2x}=\sum\limits_{k=0}^n 2^kx^k+o(x^n).$$
${\bf 5. \frac{1}{3x+4}}$
Solution.
$$\frac{1}{3x+4}=\frac{1}{4}\frac{1}{\frac{3}{4}x+1}=\frac{1}{4}\sum\limits_{k=0}^n(-1)^k\left(\frac{3}{4}\right)^kx^k+o(x^n).$$
Computing limits using Taylor formula.
Let's find $\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)},$ where $f(0)=g(0)=0.$ Assuming that the functions $f(x)$ and $g(x)$ can be expanded using the Maclaurin formula, we will limit ourselves to the first non-zero terms in the expansion of these functions:
$f(x)=ax^n+o(x^n),$ $a\neq 0,$ $g(x)=bx^m+o(x^m),$ $b\neq 0.$
If $m=n,$ then
$$\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)}=\lim\limits_{x\rightarrow 0}\frac{ax^n+o(x^n)}{bx^n+o(x^n)}=\frac{a}{b};$$
If $n>m$ , then $$\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)}=0;$$
if $m>n$ ,then $$\lim\limits_{x\rightarrow 0}\frac{f(x)}{g(x)}=\infty.$$
Examples.
Compute the limits using the Taylor formula.
1. $\lim\limits_{x\rightarrow 0}\frac{\ln (1+x)-x}{x^2}.$
Solution.
$$\lim\limits_{x\rightarrow 0}\frac{\ln (1+x)-x}{x^2}=\lim\limits_{x\rightarrow 0}\frac{x-\frac{x^2}{2}-x}{x^2}=-\frac{1}{2}.$$
2. $\lim\limits_{x\rightarrow 0}\frac{e^x-1-x}{x^2}.$
Solution.
$$\lim\limits_{x\rightarrow 0}\frac{e^x-1-x}{x^2}=\lim\limits_{x\rightarrow 0}\frac{1+x+\frac{x^2}{2}-1-x}{x^2}=\frac{1}{2}.$$
3. $\lim\limits_{x\rightarrow 0}\frac{\cos x-1+\frac{x^2}{2}}{x^4}.$
Solution.
$$\lim\limits_{x\rightarrow 0}\frac{\cos x-1+\frac{x^2}{2}}{x^4}=\lim\limits_{x\rightarrow 0}\frac{1-\frac{x^2}{2}+\frac{x^4}{4!}-1+\frac{x^2}{2}}{x^4}=\frac{1}{24}.$$
4. $\lim\limits_{x\rightarrow 0}\frac{tg x-\sin x}{x^3}.$
Solution.
$$\lim\limits_{x\rightarrow 0}\frac{tg x-\sin x}{x^3}=\lim\limits_{x\rightarrow 0}\frac{x+\frac{x^3}{3}-x+\frac{x^3}{3}}{x^3}=\frac{2}{3}.$$
5. $\lim\limits_{x\rightarrow 0}\frac{\sqrt{1+2tg x}-e^x+x^2}{arctg x-\sin x}.$
Solution.
Let's expand the numerator using the Taylor formula:
$tg x=x+\frac{x^3}{3}+o(x^3),\,\, x\rightarrow 0$
$2tg x=2x+\frac{2x^3}{3}+o(x^3),\,\, x\rightarrow 0$
$\sqrt{1+t}=(1+t)^{\frac{1}{2}}=1+\frac{1}{2}t-\frac{1}{8}t^2+\frac{1}{16}t^3+o(t^3),\,\, t\rightarrow 0;$
Thus,
$$\sqrt{1+2tg x}=1+\frac{1}{2}2 tg x-\frac{1}{8}(2 tg x)^2+\frac{1}{16}(2 tg x)^3+o(tg^3x)=$$
$$=1+tg x-\frac{1}{2}tg^2 x+\frac{1}{2}tg^3+o(tg^3x)=$$
$$=1+x+\frac{x^3}{3}-\frac{1}{2}x^2+\frac{x^3}{2}+o(x^3)=1+x-\frac{1}{2}x^2+\frac{5}{6}x^3+o(x^3).$$
Given that $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3)$, using the Maclaurin formula, we find the numerator of the fraction as
$$\sqrt{1+2tg x}-e^x+x^2=$$ $$=1+x-\frac{1}{2}x^2+\frac{5}{6}x^3-1-x-\frac{x^2}{2}-\frac{x^3}{6}+x^2+o(x^3)=$$ $$=\frac{2}{3}x^3+o(x^3),\,\,\,x\rightarrow 0.$$
Next, we expand the denominator:
$\sin x= x-\frac{x^3}{6}+o(x^3);$
$\arcsin x=x+\frac{x^3}{6}+o(x^3).$
Hence, $\arcsin x-\sin x=\frac{x^3}{3}+o(x^3).$
Thus, the fraction can be represented as $$\frac{\frac{2}{3}x^3+o(x^3)}{\frac{1}{3}x^3+o(x^3)}.$$
From here
$$\lim\limits_{x\rightarrow 0}\frac{\sqrt{1+2tg x}-e^x+x^2}{arctg x-\sin x}=2.$$
Answer: 2.
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