Chủ đề này có 3 trả lời

[minhson95]

tính:

$\lim_{x \rightarrow -\propto} (\sqrt{x^2+1}+\sqrt[3]{x^3-1})$

[okpro]

=$\lim_{n \to -\infty}$(-$x\sqrt{1+\dfrac{1}{x^2}}$+$x\sqrt[3]{1-\dfrac{1}{x^3}}$)=0

[phuonganh_lms]

=$\lim_{n \to -\infty}$(-$x\sqrt{1+\dfrac{1}{x^2}}$+$x\sqrt[3]{1-\dfrac{1}{x^3}}$)=0

Sai rồi, nếu đặt $-x$ ra ngoài thì sẽ có dạng $-\infty.0$

[nucnt772]

$\lim_{x\rightarrow-\infty}$($\sqrt{x^{2}+1}+\sqrt[3]{x^{3}-1}$)

= $\lim_{x\rightarrow-\infty}$$\sqrt{x^{2}+1}+x-x+\sqrt[3]{x^{3}-1}$

= $\lim_{x\rightarrow-\infty}$[($\sqrt{x^{2}+1}+x$)+($\sqrt[3]{x^{3}-1}-x$)]

= $\lim_{x\rightarrow-\infty}$[$\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}$+$\frac{x^{3}-1-x^{3}}{\sqrt[3]{(x^{3}-1)^{2}}+x.\sqrt[3]{x^{3}-1}+x^{2}}$]

= $\lim_{x\rightarrow-\infty}$[$\frac{1}{\sqrt{x^{2}+1}-x}$+$\frac{-1}{\sqrt[3]{(x^{3}-1)^{2}}+x.\sqrt[3]{x^{3}-1}+x^{2}}$]

= $\lim_{x\rightarrow-\infty}$[$\frac{1}{-x.\sqrt{1+\frac{1}{x^{2}}}-x}$+$\frac{-1}{x^{2}.\sqrt[3]{(1-\frac{1}{x^{3}})^{2}}+x^{2}.\sqrt[3]{1-\frac{1}{x^{3}}}+x^{2}}$

= $\lim_{x\rightarrow-\infty}$[$\frac{\frac{1}{-x}}{\sqrt{1+\frac{1}{x^{2}}}+1}$+$\frac{\frac{-1}{x^{2}}}{\sqrt[3]{(1-\frac{1}{x^{3}})^{2}}+\sqrt[3]{(1-\frac{1}{x^{3}})}+1}$]

= 0