Chủ đề này có 4 trả lời

[DOTOANNANG]

Chứng minh rằng : 

$$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= 1$$

Spoiler

$\sin \mathit{x}= \text{A}\left [ ...\,\left ( \mathit{x}+ \mathit{2}\,\pi \right )\left ( \mathit{x}+ \pi \right )x\left ( \mathit{x}- \pi \right )\left ( \mathit{x}- \mathit{2}\,\pi \right )\,... \right ]$ 

$\text{A}^{\,-\,\mathit{1}}= \frac{\pi }{\mathit{2}}\,{\prod\limits_{\mathit{k}= \mathit{1}}^{\infty }}\left [ \left ( \frac{\pi}{\mathit{2}} \right )^{\,\mathit{2}}- \mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right ]$

Dùng nhân tử Wallis : $\lceil$ https://en.wikipedia.../Wallis_product $\rfloor$ , ta được :

$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= \text{A}\left [ ...\,\left ( \mathit{2}\,\pi \right )\,\pi \,\left ( -\,\pi \right )\,\left ( -\,\mathit{2}\,\pi \right )\,... \right ]= \text{A}\,\prod\limits_{k= 1}^{\infty } \left ( -\,\mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right )= \mathit{1}$

[DOTOANNANG]

 

[DOTOANNANG]

với ${\it a}= \left [ \frac{\pi}{{\it 6}},\,\frac{\pi}{{\it 3}},\,...,\,\frac{{\it 5}\,\pi}{{\it 6}} \right ]$ trong công thức $\lceil$ https://www.desmos.c...ator/ldwe8ai6iq $\rfloor$ gồm $\it{2}$ đồ thị $y= \tan\,(\,t\,)\,x$ và $y= \tan\,(\,a\,)\,t\,x$.

[chanhquocnghiem]

Chứng minh rằng : 

$$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= 1$$

Spoiler

$\sin \mathit{x}= \text{A}\left [ ...\,\left ( \mathit{x}+ \mathit{2}\,\pi \right )\left ( \mathit{x}+ \pi \right )x\left ( \mathit{x}- \pi \right )\left ( \mathit{x}- \mathit{2}\,\pi \right )\,... \right ]$ 

$\text{A}^{\,-\,\mathit{1}}= \frac{\pi }{\mathit{2}}\,{\prod\limits_{\mathit{k}= \mathit{1}}^{\infty }}\left [ \left ( \frac{\pi}{\mathit{2}} \right )^{\,\mathit{2}}- \mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right ]$

Dùng nhân tử Wallis : $\lceil$ https://en.wikipedia.../Wallis_product $\rfloor$ , ta được :

$\lim_{\mathit{x}\,\rightarrow \,\mathit{0}}\frac{\sin \mathit{x}}{\mathit{x}}= \text{A}\left [ ...\,\left ( \mathit{2}\,\pi \right )\,\pi \,\left ( -\,\pi \right )\,\left ( -\,\mathit{2}\,\pi \right )\,... \right ]= \text{A}\,\prod\limits_{k= 1}^{\infty } \left ( -\,\mathit{k}^{\,\mathit{2}}\,\pi ^{\,\mathit{2}} \right )= \mathit{1}$

$\lim_{x\to0}\frac{\sin x}{x}=\lim_{x\to0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}{x}=\lim_{x\to0}\left ( 1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+... \right )=1$

[DOTOANNANG]

$$O(\,x^{\,6}\,)= -\,\frac{x^{\,7}}{7\,!}+ ... $$