Các bạn tính giúp mình mấy câu này với:
1. \(\lim\limits_{x\rightarrow\left(-1\right)-}\dfrac{\sqrt{x^2-3x-4}}{1-x^2}\)
2. \(\lim\limits_{x\rightarrow2^+}\left(\dfrac{1}{x-2}-\dfrac{x+1}{\sqrt{x+2}-2}\right)\)
3. \(\lim\limits_{x\rightarrow+\infty}\dfrac{3x^2-5sin2x+7cos^2x}{2x^2+2}\)
4. \(\lim\limits_{x\rightarrow+\infty}\left(x.sin\left(\dfrac{1}{3x}\right)\right)\)
5. \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{2x+1}.\sqrt[3]{3x+1}.\sqrt[4]{4x+1}-1}{x}\)
6. \(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{9x+4}-\sqrt[3]{4x^{^2}+8}}{sinx}\right)\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
3.
\(\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7\cos ^2x}{2x^2+2}=\lim\limits_{x\to +\infty}\frac{3x^2-5\sin 2x+7(1-\sin ^2x)}{2x^2+2}\)
\(=\lim\limits_{x\to +\infty}\frac{3(x^2+1)-5\sin 2x+4-7\sin ^2x}{2x^2+2}\)
\(=\lim\limits_{x\to +\infty}\left[\frac{3}{2}-5.\frac{\sin 2x}{2x}.\frac{2x}{2x^2+2}+\frac{2}{x^2+1}-7.(\frac{\sin x}{x})^2.\frac{x^2}{2x^2+2}\right]\)
\(=\frac{3}{2}-5.0.0+0-7.0.\frac{1}{2}=\frac{3}{2}\) (nhớ rằng \(\lim\limits_{t\to \infty}\frac{\sin t}{t}=0\))
4.
Với $x\to +\infty$ thì $\frac{1}{3x}\to 0$
Nhớ rằng: $\lim\limits_{t\to 0}\frac{\sin t}{t}=1$ nên:
\(\lim\limits_{x\to +\infty}x.\sin \frac{1}{3x}=\lim\limits_{x\to +\infty}\frac{1}{3}.\frac{\sin \frac{1}{3x}}{\frac{1}{3x}}\)
\(=\frac{1}{3}.1=\frac{1}{3}\)
5.
\(\lim\limits_{x\to 0}\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}.\sqrt[4]{4x+1}-1}{x}=\lim\limits_{x\to 0}\frac{\sqrt{2x+1}.\sqrt[3]{3x+1}(\sqrt[4]{4x+1}-1)+\sqrt{2x+1}(\sqrt[3]{3x+1}-1)+(\sqrt{2x+1}-1)}{x}\)
\(=\lim\limits_{x\to 0}\left[\sqrt{2x+1}.\sqrt[3]{3x+1}.\frac{4}{\sqrt[4]{(4x+1)^3}+\sqrt[4]{(4x+1)^2}+\sqrt[4]{4x+1}+1}+\sqrt{2x+1}.\frac{3}{\sqrt[3]{(3x+1)^2}+\sqrt[3]{3x+1}+1}+\frac{2}{\sqrt{2x+1}+1}\right]\)
\(=3\)
6.
\(\lim\limits_{x\to 0}\left(\frac{\sqrt{9x+4}-\sqrt[3]{4x^2+8}}{\sin x}\right)=\lim\limits_{x\to 0}\frac{\sqrt{9x+4}-2-(\sqrt[3]{4x^2+8}-2)}{\sin x}\)
\(=\lim\limits_{x\to 0}\frac{\frac{9x}{\sqrt{9x+4}+2}-\frac{4x^2}{\sqrt[3]{(4x^2+8)^2}+2\sqrt[3]{4x^2+8}+4}}{\sin x}\)
\(=\lim\limits_{x\to 0}(\frac{9}{\sqrt{9x+4}+2}-\frac{4x}{\sqrt[3]{(4x^2+8)^2}+2\sqrt[3]{4x^2+8}+4}).\frac{x}{\sin x}\)
\(=\frac{9}{4}\) do:
\(\lim\limits_{x\to 0}\frac{x}{\sin x}=1\) và \(\lim\limits_{x\to 0}(\frac{9}{\sqrt{9x+4}+2}-\frac{4x}{\sqrt[3]{(4x^2+8)^2}+2\sqrt[3]{4x^2+8}+4})=\frac{9}{4}-0=\frac{9}{4}\)
