Kiểm tra lại đề bài, giới hạn này không tồn tại

sao có GP lại ko có huy hiệu hỏi thôi

Thấy : \(\sqrt{x^2+x+3}-x^2+1=\sqrt{x^2+x+3}-\left(x^2-1\right)=\dfrac{x^2+x+3-\left(x^2-1\right)^2}{\sqrt{x^2+x+3}+x^2-1}\)

\(=\dfrac{x^2+x+3-x^4+2x^2-1}{...}=\dfrac{-x^4+3x^2+x+2}{...}\)

\(=\dfrac{-\left(x-2\right)\left(x^3+2x^2+x+1\right)}{...}\)

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

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\dfrac{-\left(2^3+2.2^2+2+1\right)}{4.\left[\sqrt{2^2+2+3}+2^2-1\right]}=-\dfrac{19}{24}\)

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{x^2+x+3}-x^2+1}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\dfrac{2x+1}{2\sqrt{x^2+x+3}}-2x}{2x}=\dfrac{\dfrac{2.2+1}{2\sqrt{4+2+3}}-4}{4}=-\dfrac{19}{24}\)

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

a/ L'Hospital:

 \(=\lim\limits_{x\rightarrow2}\dfrac{x-\left(x+2\right)^{\dfrac{1}{2}}}{\left(4x+1\right)^{\dfrac{1}{2}}-3}=\lim\limits_{x\rightarrow2}\dfrac{1-\dfrac{1}{2}\left(x+2\right)^{-\dfrac{1}{2}}}{\dfrac{1}{2}\left(4x+1\right)^{-\dfrac{1}{2}}.4}=\dfrac{1-\dfrac{1}{2}.4^{-\dfrac{1}{2}}}{2.9^{-\dfrac{1}{2}}}=\dfrac{9}{8}\)

b/ L'Hospital:\(=\lim\limits_{x\rightarrow1}\dfrac{\left(2x+7\right)^{\dfrac{1}{2}}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{2}\left(2x+7\right)^{-\dfrac{1}{2}}.2+1}{3x^2-8x}=\dfrac{9^{-\dfrac{1}{2}}+1}{3-8}=-\dfrac{4}{15}\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{7+x^3}-\sqrt{3+x^2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[3]{7+x^3}-2\right)-\left(\sqrt{3+x^2}-2\right)}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^3-1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x^2-1}{\sqrt{3+x^2}+2}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{x^2+x+1}{\left(\sqrt[3]{7+x^3}\right)^2+2\sqrt[3]{7+x^3}+4}-\dfrac{x+1}{\sqrt{3+x^2}+2}}{1}=\dfrac{3}{12}-\dfrac{2}{4}=\dfrac{1}{4}-\dfrac{1}{2}=-\dfrac{1}{4}\).

Ta xét:

\(\sqrt{1+2x}\cdot\sqrt[3]{1+3x}-1\)

\(=\sqrt{1+2x}-\sqrt{1+2x}+\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1\)

\(=\left(\sqrt{1+2x}-1\right)+\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)\)

Xét giới hạn trên:

\(\Rightarrow^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\)

   \(=^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)+^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\left(\sqrt[3]{1+2x}-1\right)}{3}\right)\)

Tính giới hạn từng thành phần:

* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}-1}{x}\right)=^{lim}_{x\rightarrow0}\left(\dfrac{1+2x-1}{x\left(\sqrt{1+2x}+1\right)}\right)\)

  \(=^{lim}_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+2x}+1}\right)=\dfrac{2}{\sqrt{1+2\cdot0}+1}=1\left(1\right)\)

* \(^{lim}_{x\rightarrow0}\left(\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}\right)\)

  \(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{1+2x-1}{x\left(\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1\right)}\right)\)

  \(=^{lim}_{x\rightarrow0}\left(\sqrt{1+2x}\cdot\dfrac{2}{\left(\sqrt[3]{1+2x}\right)^2+\sqrt[3]{1+2x}+1}\right)\)

  \(=\sqrt{1+2\cdot0}\cdot\dfrac{2}{(\sqrt[3]{1+2\cdot0})^2+\sqrt[3]{1+2\cdot0}+1}\)

  \(=\dfrac{2}{3}\left(2\right)\)

Lấy \(\left(1\right)+\left(2\right)\) ta được:

\(^{lim}_{x\rightarrow0}\dfrac{\sqrt{1+2x}\cdot\sqrt[3]{1+2x}-1}{x}=1+\dfrac{2}{3}=\dfrac{5}{3}\)

a: \(=lim_{x->-\infty}\dfrac{2x-5+\dfrac{1}{x^2}}{7-\dfrac{1}{x}+\dfrac{4}{x^2}}\)

\(=\dfrac{2x-5}{7}\)

\(=\dfrac{2}{7}x-\dfrac{5}{7}\)

\(=-\infty\)

b: \(=lim_{x->+\infty}x\sqrt{\dfrac{1+\dfrac{1}{x}+\dfrac{3}{x^2}}{3x^2+4-\dfrac{5}{x^2}}}\)

\(=lim_{x->+\infty}x\sqrt{\dfrac{1}{3x^2+4}}=+\infty\)