tính lim \(\dfrac{cosx}{x}\) và lim \(\dfrac{sin^2x}{x}\)khi x-> 0
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Tính các giới hạn
a) \(\lim\limits_{x\rightarrow a}\dfrac{\sin x-\sin a}{x-a}\)
b) \(\lim\limits_{x\rightarrow1}\left(1-x\right)\tan\dfrac{\pi x}{2}\)
c) \(\lim\limits_{x\rightarrow\dfrac{\pi}{3}}\dfrac{2\sin^2x+\sin x-1}{2\sin^2x-3\sin x+1}\)
d) \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
tính giới hạn sau:
\(\lim\limits_{x\rightarrow\dfrac{\pi}{6}}\dfrac{\sqrt{3}sinx-cosx}{sin\left(\dfrac{\pi}{3}-2x\right)}\)
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 \(x\to \frac{\pi}{6}\Rightarrow \frac{\pi}{6}-x\to 0\) )
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Tính \(\lim\limits_{x\rightarrow a}\dfrac{\sin^2x-\sin^2a}{x^2-a^2}\)
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Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
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Tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x-\tan x}{\left(\sqrt{1+2x}-\sqrt[3]{1+3x}\right)\cdot\sin10x}\)
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Tính: \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{sinx}{cosx}-sinx}{sin^3x}=\lim\limits_{x\rightarrow0}\dfrac{1-cosx}{cosx.sin^2x}=\lim\limits_{x\rightarrow0}\dfrac{2sin^2\dfrac{x}{2}}{4cosx.cos^2\dfrac{x}{2}sin^2\dfrac{x}{2}}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{1}{2cosx.cos^2\dfrac{x}{2}}=\dfrac{1}{2}\)
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Tính \(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sin x}-\dfrac{3}{\sin3x}\right)\dfrac{1}{x}\)
Tính giới hạn
a) \(\lim\limits_{x->0}\dfrac{\sqrt[m]{2x+1}-1}{\sqrt[n]{x+1}-1}\)
b) \(\lim\limits_{x->3}\dfrac{\sqrt[4]{5x+1}-2}{x-3}\)
Tính \(\lim\limits_{x\rightarrow0}\dfrac{\sin x\sin2x...\sin nx}{x^n}\).