Tính giá trị giới hạn lim (x → 0) \(\dfrac{\left(x^2+\Pi^{21}\right)\sqrt[7]{1-2x}-\Pi^{21}}{x}\) là:
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HT
18 tháng 2 2021
\(=\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+\pi^{21}\right)\left(1-2x\right)^{\dfrac{1}{7}}-\pi^{21}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{7}\left(1-2x\right)^{-\dfrac{6}{7}}.\left(-2\right)\left(x^2+\pi^{21}\right)+2x\left(1-2x\right)^{\dfrac{1}{7}}}{1}\)
\(=\dfrac{1}{7}.\left(-2\right).\pi^{21}=...\)
AH
Akai Haruma
Giáo viên
11 tháng 3 2018
Lời giải:
Ta có:
Áp dụng công thức lượng giác: \(\sin (a-b)=\sin a\cos b-\cos a\sin b\)
thì:
\(\sqrt{3}\sin x-\cos x=-2\left(\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x\right)=-2\left(\sin \frac{\pi}{6}\cos x-\cos \frac{\pi}{6}\sin x\right)\)
\(=-2\sin \left(\frac{\pi}{6}-x\right)\)
Do đó: \(\lim_{x\to \frac{\pi}{6}}\frac{\sqrt{3}\sin x-\cos x}{\sin (\frac{\pi}{3}-2x)}=-2\lim_{x\to \frac{\pi}{6}}\frac{\sin \left ( \frac{\pi}{6}-x \right )}{\sin \left [ 2(\frac{\pi}{6}-x) \right ]}\)
\(=-\lim_{x\to \frac{\pi}{6}}\frac{\sin \left ( \frac{\pi}{6}-x \right )}{\frac{\pi}{6}-x}.\lim_{x\to \frac{\pi}{6}}\frac{1}{\frac{\sin\left [ 2(\frac{\pi}{6}-x) \right ]}{2(\frac{\pi}{6}-x)}}=-1.1.1=-1\)
(sử dụng công thức \(\lim_{t\to 0} \frac{\sin t}{t}=1\) . Trong TH bài toán \(x\to \frac{\pi}{6}\Rightarrow \frac{\pi}{6}-x\to 0\) )
MA
3
HP
19 tháng 3 2022
Đặt \(t=x-\dfrac{\pi}{4}\), khi đó:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sqrt{2}cosx-1}{\sqrt{2}sinx-1}=\lim\limits_{t\rightarrow0}\dfrac{\sqrt{2}cos\left(t+\dfrac{\pi}{4}\right)-1}{\sqrt{2}sin\left(t+\dfrac{\pi}{4}\right)-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{cost-sint-1}{cost+sint-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{1-2sin^2\dfrac{t}{2}-2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}{1-2sin^2\dfrac{t}{2}+2sin\dfrac{t}{2}.cos\dfrac{t}{2}-1}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}+cos\dfrac{t}{2}\right)}{-2sin\dfrac{t}{2}\left(sin\dfrac{t}{2}-cos\dfrac{t}{2}\right)}\)
\(=\lim\limits_{t\rightarrow0}\dfrac{sin\dfrac{t}{2}+cos\dfrac{t}{2}}{sin\dfrac{t}{2}-cos\dfrac{t}{2}}\)
\(=-1\)
KS
30 tháng 3 2022
Cho hàm số y=f(x)y=f(x) có đạo hàm và liên tục trên [0;π2][0;π2]thoả mãn f(x)=f′(x)−2cosxf(x)=f′(x)−2cosx. Biết f(π2)=1f(π2)=1, tính giá trị f(π3)f(π3)
A. √3+1/2 B. √3−1/2 C. 1−√3/2 D. 0
HT
18 tháng 2 2021
a/ \(=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}-x}+\lim\limits_{x\rightarrow-\infty}\dfrac{3x^3-1-x^3}{\sqrt[3]{\left(3x^3-1\right)^2}+x\sqrt[3]{3x^3-1}+x^2}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{\dfrac{1}{x}}{-\sqrt{\dfrac{x^2}{x^2}+\dfrac{1}{x^2}}-\dfrac{x}{x}}+\lim\limits_{x\rightarrow-\infty}\dfrac{-\dfrac{1}{x^2}}{\dfrac{\sqrt[3]{\left(3x^3-1\right)^2}}{x^2}+\dfrac{x\sqrt[3]{3x^3-1}}{x^2}+\dfrac{x^2}{x^2}}=0\)
b/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+x-x^2}{\sqrt{x^2+x}+x}+\lim\limits_{x\rightarrow+\infty}\dfrac{x^3-x^3+x^2}{x^2+x\sqrt[3]{x^3-x^2}+\sqrt[3]{\left(x^3-x^2\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x}{x}}{\sqrt{\dfrac{x^2}{x^2}+\dfrac{x}{x^2}}+\dfrac{x}{x}}+\lim\limits_{x\rightarrow+\infty}\dfrac{\dfrac{x^2}{x^2}}{\dfrac{x^2}{x^2}+\dfrac{x\sqrt[3]{x^3-x^2}}{x^2}+\dfrac{\sqrt[3]{\left(x^3-x^2\right)^2}}{x^2}}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
c/ \(=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-1-2x-1}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{4x^2-1}+\sqrt[3]{\left(2x+1\right)^2}}\)
\(=\lim\limits_{x\rightarrow+\infty}\dfrac{-\dfrac{2}{x^{\dfrac{2}{3}}}}{\dfrac{\sqrt[3]{\left(2x-1\right)^2}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{4x^2-1}}{x^{\dfrac{2}{3}}}+\dfrac{\sqrt[3]{\left(2x+1\right)^2}}{x^{\dfrac{2}{3}}}}=0\)
Check lai ho minh nhe :v
