Câu 1.

\(lim\left(\sqrt{n^2+2n+5}-\sqrt{n^2+n}\right)\)

Nhân liên hợp ta đc:

\(lim\left(\dfrac{n^2+2n+5-\left(n^2+n\right)}{\sqrt{n^2+2n+5}+\sqrt{n^2+n}}\right)\)

\(=lim\left(\dfrac{n+5}{n\sqrt{1+\dfrac{2}{n}+\dfrac{5}{n^2}}+n\sqrt{1+\dfrac{1}{n}}}\right)\)

\(=lim\left(\dfrac{n+\dfrac{5}{n}}{n\sqrt{1}+n\sqrt{1}}\right)=lim\left(\dfrac{n}{2n}\right)=lim\dfrac{1}{2}=\dfrac{1}{2}\)

Chọn C

\(lim\left(x-\sqrt{x^2-2x}\right)\)

\(=lim\left(x-x\sqrt{1-\dfrac{2}{x}}\right)\)

\(=lim\left[x\left(1-\sqrt{1-\dfrac{2}{x}}\right)\right]\)

\(=0\)

x dần tới mấy bạn.

a.

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4+x}-2}{4x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{4+x}-2\right)\left(\sqrt{4+x}+2\right)}{4x\left(\sqrt{4+x}+2\right)}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{x}{4x\left(\sqrt{4+x}+2\right)}=\lim\limits_{x\rightarrow0}\dfrac{1}{4\left(\sqrt{4+x}+2\right)}=\dfrac{1}{4\left(\sqrt{4+0}+2\right)}=\dfrac{1}{16}\)

b.

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{x+7}-2\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}{\left(x-1\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}\)

\(=\dfrac{1}{\sqrt[3]{8^2}+2\sqrt[3]{8}+4}=\dfrac{1}{12}\)

1a.

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{4x^2-x+2}+2x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{-x+2}{\sqrt{4x^2-x+2}-2x}\)

\(=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(-1+\dfrac{2}{x}\right)}{\left|x\right|\sqrt{4-\dfrac{1}{x}+\dfrac{2}{x^2}}-2x}=\lim\limits_{x\rightarrow-\infty}\dfrac{x\left(-1+\dfrac{2}{x}\right)}{x\left(-\sqrt{4-\dfrac{1}{x}+\dfrac{2}{x^2}}-2\right)}\)

\(=\dfrac{-1+0}{-\sqrt{4-0+0}-2}=\dfrac{1}{4}\)

b.

\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+2x-3}-x\right)=\lim\limits_{x\rightarrow+\infty}\dfrac{2x-3}{\sqrt{x^2+2x-3}+x}\)

\(=\lim\limits_{x\rightarrow+\infty}\dfrac{x\left(2-\dfrac{3}{x}\right)}{x\left(\sqrt{1+\dfrac{2}{x}-\dfrac{3}{x^2}}+1\right)}=\dfrac{2-0}{\sqrt{1+0-0}+1}=1\)

2a.

\(\lim\limits_{x\rightarrow-\infty}\left(-x^3+x^2-x+1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)\)

Mà \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x^3=-\infty\\\lim\limits_{x\rightarrow-\infty}\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=-1< 0\end{matrix}\right.\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x^3\left(-1+\dfrac{1}{x}-\dfrac{1}{x^2}+\dfrac{1}{x^3}\right)=+\infty\)

b.

\(\lim\limits_{x\rightarrow-\infty}\left(2x-\sqrt{4x^2+2x-1}\right)=\lim\limits_{x\rightarrow-\infty}\left(2x-\left|x\right|\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)\)

\(=\lim\limits_{x\rightarrow-\infty}\left(2x+x\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=\lim\limits_{x\rightarrow-\infty}x\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)\)

Do \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow-\infty}x=-\infty\\\lim\limits_{x\rightarrow-\infty}\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=2+2=4>0\end{matrix}\right.\)

\(\Rightarrow\lim\limits_{x\rightarrow-\infty}x\left(2+\sqrt{4+\dfrac{2}{x}-\dfrac{1}{x^2}}\right)=-\infty\)

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