D
duynhan1
[TEX]\huge \lim_{x%20\to%200}x\sin%20\frac{1}{x} = \lim_{x%20\to%200}x^2 \frac{\sin%20\frac{1}{x} }{\frac1x} = 0[/TEX]
[TEX]\huge \lim_{x%20\to%200}x\sin%20\frac{1}{x} = \lim_{x%20\to%200}x^2 \frac{\sin%20\frac{1}{x} }{\frac1x} = 0[/TEX]
B
bigbang195
N
ngomaithuy93
[TEX] A=cos.\frac{a}{2^n}cos.\frac{a}{2^{n-1}}...cos.\frac{a}{2^2}cos.\frac{a}{2}[/TEX]})
[TEX] \Leftrightarrow 2A.sin.\frac{a}{2^n}=sin.\frac{a}{2^{n-1}}. ... . cos.\frac{a}{2^2}cos.\frac{a}{2}[/TEX]
[TEX] \Leftrightarrow A=\frac{sina}{2sin.\frac{a}{2^n}}[/TEX]
[TEX] \lim_{n\to \infty} A=0 [/TEX]
V
vivietnam
C1
dùng L' là ra
C2
[TEX]=\lim_{x \to 3}(\frac{x^2-9}{sin(\frac{\pi}{2}-\frac{\pi.x}{6})})=\lim_{x \to 3}\frac{x^2-9}{\frac{\pi}{2}-\frac{\pi.x}{6}}.\frac{\frac{\pi}{2}-\frac{\pi.x}{6}}{sin(\frac{\pi}{2}-\frac{\pi.x}{6})}=\lim_{x \to 3}\frac{x+3}{\frac{\pi}{6}}=\frac{36}{\pi}[/TEX]
tưởng mất nick này
hú hồn
Last edited by a moderator:
V
vivietnam
tương tự bài hôm trước
[TEX]=\lim_{\frac{1}{x} \to 0}(-2cos(\frac{\sqrt{x+1}+\sqrt{x}}{2}).sin(\frac{\sqrt{x+1}-\sqrt{x}}{2})=\lim_{\frac{1}{x} \to 0}(-2.cos(\frac{\sqrt{x+1}+\sqrt{x}}{2}).sin(\frac{1}{2.(\sqrt{x+1}+\sqrt{x})})[/TEX]
ta có [TEX]\lim_{\frac{1}{x} \to 0}(\frac{1}{2.(\sqrt{x+1}+\sqrt{x})})=0[/TEX]
mà [TEX]|cos(\frac{\sqrt{x+1}+\sqrt{x}}{2})| \leq1[/TEX]
\Rightarrow[TEX]\lim_{\frac{1}{x} \to 0}(sin{\sqrt{x+1}}-sin\sqrt{x})=0[/TEX]
Last edited by a moderator:
V
vivietnam
[TEX]=\lim_{x \to 0}(\frac{\sqrt{1+tanx}-1+1-\sqrt{1+sinx}}{x^3}=\lim_{x \to 0}(\frac{tanx-sinx}{2x^3}=\lim_{x \to 0}(\frac{1-cosx}{2cosx.x^2}=\lim_{x \to 0}(\frac{2.sin^2{\frac{x}{2}}}{2x^2})=4[/TEX]
dùng cách nhân liên hợp cũng ra
[TEX]=\lim_{\frac{1}{x} \to 0}(\frac{(\sqrt{x+1}+\sqrt{x}).sin(2.(\sqrt{x+1}+\sqrt{x}))}{2x.sin(\sqrt{x+1}+\sqrt{x})}[/TEX]})
bằng cách đặt [TEX]\frac{1}{x}=t [/TEX]
\Rightarrowlim =0
Last edited by a moderator:
B
bigbang195
V
vivietnam12345
em sử dụng giới hạn
[TEX]\lim_{x \to x_0}\frac{sinu(x)}{u(x)}=1[/TEX]
thêm vào tử với mẫu là được cái đấy
thế nào mà cứ quên mất cái mật khẩu của nick kia
V
vivietnam
V
vivietnam
câu này hay-\sin%20(\sin%20x)}{\tan%20x-\sin%20x}})
mọi người làm rồi đưa cách giải lên đi
nếu học rùi thì có thể dùng L' sẽ ra nhanh
còn chưa học thì dùng các phép biến đổi lượng giác và giới hạn cơ bản là được
L
letrang3003
[TEX]\huge1) \ \blue{ \lim_{x \to 0} x\cos \frac{1}{x}[/TEX]
[TEX]\huge2) \ \blue{\lim_{x \to 0} \frac{x}{a} \bigg[\frac{b}{x} \bigg] (a,b>0)[/TEX]
[TEX]\huge3) \ \blue{\huge \lim_{x \to +\infty} x(\sqrt{x^2+1}-\sqrt[3]{x^3+1})[/TEX]
[TEX]\huge4) \ \blue{\huge\lim_{x \to 0} x \bigg[\frac{1}{x}\bigg][/TEX]
[TEX]\huge5) \ \blue{\huge\lim_{ x \to 0} \frac{[x]}{x}[/TEX]
[TEX]\huge6) \ \blue{\huge\lim_{x \to 0} \frac{cos \bigg(\frac{\pi}{2} \cos x\bigg)}{\sin x(\sin x)}[/TEX]
[TEX]\huge2) \ \blue{\lim_{x \to 0} \frac{x}{a} \bigg[\frac{b}{x} \bigg] (a,b>0)[/TEX]
[TEX]\huge3) \ \blue{\huge \lim_{x \to +\infty} x(\sqrt{x^2+1}-\sqrt[3]{x^3+1})[/TEX]
[TEX]\huge4) \ \blue{\huge\lim_{x \to 0} x \bigg[\frac{1}{x}\bigg][/TEX]
[TEX]\huge5) \ \blue{\huge\lim_{ x \to 0} \frac{[x]}{x}[/TEX]
[TEX]\huge6) \ \blue{\huge\lim_{x \to 0} \frac{cos \bigg(\frac{\pi}{2} \cos x\bigg)}{\sin x(\sin x)}[/TEX]
B
bigbang195
[TEX]\huge \lim_{x \to +\infty} x(\sqrt{x^2+1}-\sqrt[3]{x^3+1}) = \lim_{x \to +\infty} \frac{x}{\sqrt{x^2+1} + x}-\frac{x}{\sqrt[3]{(x^3+1)^2}+x\sqrt[3]{x^3+1}+x^2} = \lim_{x \to +\infty} \frac{1}{2} - \frac{0}{3} = \frac12[/TEX]
Last edited by a moderator:
L
longnhi905
sai rồi sinx~x khi x ->0 mà kia là 1/x mà vậy không thể dùng ~. mà 1/x -> vô cùng, x->0 vậy kết quả là 0
Last edited by a moderator:
D
duynhan1
[TEX]\huge \red \fbox{A_1 = \lim_{x\to 2} (\frac{1}{x-2} - \frac{12}{x^3-8})[/TEX]
[TEX]\huge \blue \fbox{A_2 = \lim_{x\to 0} (\frac{\sqrt{1-2x+x^3} - 1 - x}{x})[/TEX]
[TEX]\huge \blue \fbox{A_2 = \lim_{x\to 0} (\frac{\sqrt{1-2x+x^3} - 1 - x}{x})[/TEX]
D
duynhan1
[TEX]\huge \red \fbox{A_3 = \lim_{x \to 1 } ( \frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3})} [/TEX]
[TEX]\huge \blue \fbox{A_4 = \lim_{x \to 1} (\frac{\sqrt[4]{x}-1}{\sqrt[3]{x}-1})} [/TEX]
[TEX]\huge \blue \fbox{A_4 = \lim_{x \to 1} (\frac{\sqrt[4]{x}-1}{\sqrt[3]{x}-1})} [/TEX]
Last edited by a moderator:
B
bigbang195
D
duynhan1
m,n [TEX]\huge \in[/TEX] N
[TEX]\huge \bold{ \fbox{A_{5*} = \lim_{x \to 1} (\frac{n}{1-x^n} - \frac{1}{1-x})[/TEX]
[TEX]\huge \bold{ \fbox{A_{6*} = \lim_{x \to 1} (\frac{m}{1-x^m} - \frac{n}{1-x^n})[/TEX]
[TEX]\huge \bold{ \fbox{A_{5*} = \lim_{x \to 1} (\frac{n}{1-x^n} - \frac{1}{1-x})[/TEX]
[TEX]\huge \bold{ \fbox{A_{6*} = \lim_{x \to 1} (\frac{m}{1-x^m} - \frac{n}{1-x^n})[/TEX]
Last edited by a moderator:
B
bigbang195
B
bigbang195
[TEX]\huge {\color{red} \fbox{A_{5*} = \lim_{x \to 1} \bigg(\frac{n}{1-x^n} - \frac{1}{1-x} \bigg)=\lim_{x \to 1}\frac{n-\displaystyle\sum_{i=0}^{n-1}x^i}{(1-x)\displaystyle\sum_{i=0}^{n-1}x^i}=\lim_{x \to 1}\frac{\displaystyle\sum^{n-1}_{i=1}(1-x^i)}{(1-x)\displaystyle\sum_{i=0}^{n-1}x^i}=\frac{\frac{n(n-1)}{2}}{n}=\frac{n-1}{2}}}[/TEX]
[TEX]\fbox{\huge {\color{blue} A_{6*} = \lim_{x \to 1}\bigg (\frac{m}{1-x^m} - \frac{n}{1-x^n}\bigg)\\\\=\lim_{x \to 1}\bigg(\frac{m}{1-x^m}-\frac{1}{1-x}\bigg)-\lim_{x \to 1}\bigg(\frac{n}{1-x^n}-\frac{1}{1-x}\bigg)\\\\=\frac{m-1}{2}-\frac{n-1}{2}=\frac{m-n}{2}}}[/TEX]
D
duynhan1
[TEX]\huge \red \fbox{ A_7= \lim_{x \to 0} (\frac{\sqrt[n]{1+ax}-\sqrt[n]{1+bx}}{x})[/TEX]
[TEX]\huge \blue \fbox{A_8 = \lim_{x \to +\infty }(\frac{(x+1)(x^2+1)....(x^n+1)}{\bigg((nx)^n+1 \bigg)^{\frac{n+1}{2}}} [/TEX]
[TEX]\huge \blue \fbox{A_8 = \lim_{x \to +\infty }(\frac{(x+1)(x^2+1)....(x^n+1)}{\bigg((nx)^n+1 \bigg)^{\frac{n+1}{2}}} [/TEX]