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 \(\lim\limits_{x\rightarrow\dfrac{\pi}{4}}\dfrac{\sin x-\cos x}{1-\tan x}\)
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 \(x\to \frac{\pi}{6}\Rightarrow \frac{\pi}{6}-x\to 0\) )
Tìm các giới hạn :
a) \(\lim\limits_{x\rightarrow1}\dfrac{x^2-5x+6}{x-2}\)
b) \(\lim\limits_{x\rightarrow\dfrac{\pi}{8}}\dfrac{\sin2x-\cos2x}{8x-\pi}\)
a/ \(\lim\limits_{x\to 1} f(x)=\frac{x^{2}-5x + 6}{x-2} \)
\(<=>\lim\limits_{x\to 1} f(x)=\dfrac{(x-3)(x-2)}{x-2} \)
<=>\(\lim\limits_{x\to 1} f(x)=x-3 \)
\(<=>\lim\limits_{x\to 1} f(x)=-2\)
1/ \(\lim\limits_{x\to 1}\) \(\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}\)
2/ \(\lim\limits_{x \to \ +\infty} \)\(x\left[\sqrt{4x^2+5}-\sqrt[3]{8x^3-1}\right]\)
3/ \(\lim\limits_{x\to 1}\)\(\dfrac{x^3-2x-1}{x^5-2x-1}\)
Giải giúp mình với ạ
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).
Tính \(\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sin x}-\dfrac{3}{\sin3x}\right)\dfrac{1}{x}\)
Tính: \(\lim\limits_{x\rightarrow0}\dfrac{\tan x-\sin x}{\sin^3x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{sinx}{cosx}-sinx}{sin^3x}=\lim\limits_{x\rightarrow0}\dfrac{1-cosx}{cosx.sin^2x}=\lim\limits_{x\rightarrow0}\dfrac{2sin^2\dfrac{x}{2}}{4cosx.cos^2\dfrac{x}{2}sin^2\dfrac{x}{2}}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{1}{2cosx.cos^2\dfrac{x}{2}}=\dfrac{1}{2}\)
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow\frac{\pi}{2}}\frac{sin\left(x-\frac{\pi}{4}\right)}{x}\)
b, \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{3x^2-4}-\sqrt{3x-2}}{x+1}\)
c,\(\lim\limits_{x\rightarrow0}x^2sin\frac{1}{2}\)
Tất cả đều ko phải dạng vô định, bạn cứ thay số vào tính thôi:
\(a=\frac{sin\left(\frac{\pi}{4}\right)}{\frac{\pi}{2}}=\frac{\sqrt{2}}{\pi}\)
\(b=\frac{\sqrt[3]{3.4-4}-\sqrt{6-2}}{3}=\frac{0}{3}=0\)
\(c=0.sin\frac{1}{2}=0\)
