Question

[Trần Minh]

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. \(\lim\limits_{x\rightarrow0}\left(\dfrac{\sqrt{9x+4}-\sqrt[3]{4x^{^2}+8}}{sinx}\right)\)

Answer 1

[Akai Haruma]

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 \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)

Answer 2

[Akai Haruma]

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 \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)