H24

\(lim_{x->-\infty}\left(\sqrt{x^2+1}+x-1\right)\\ lim\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)

NT
24 tháng 11 2023 lúc 20:33

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

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

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

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

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

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

b: \(\lim\limits\dfrac{\sqrt{4n^2+n-1}+n}{\sqrt{n^4+2n^3-1}-n}\)

\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\cdot\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-n^2\cdot\dfrac{1}{n}}\)

\(=\lim\limits\dfrac{n\left(\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1\right)}{n^2\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)

\(=\lim\limits\dfrac{\sqrt{4+\dfrac{1}{n}-\dfrac{1}{n^2}}+1}{n\left(\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}\right)}\)

\(=\lim\limits\dfrac{\sqrt{\dfrac{4}{n^2}+\dfrac{1}{n^3}-\dfrac{1}{n^4}}+\dfrac{1}{n}}{\sqrt{1+\dfrac{2}{n}-\dfrac{1}{n^4}}-\dfrac{1}{n}}\)

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

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