Lời giải:
\(\lim\limits _{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim\limits _{x\to 1}\frac{x-1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}.\frac{\sqrt{x}+1}{x-1}=\lim\limits _{x\to 1}\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\frac{2}{3}\)
Lời giải:
\(\lim\limits _{x\to 1}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}=\lim\limits _{x\to 1}\frac{x-1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}.\frac{\sqrt{x}+1}{x-1}=\lim\limits _{x\to 1}\frac{\sqrt{x}+1}{\sqrt[3]{x^2}+\sqrt[3]{x}+1}=\frac{2}{3}\)
Giup minh voi thanks
Lim\(\frac{x}{\sqrt{1+x}-1}\)(x=>0)
\(\lim\limits_{x\rightarrow+\infty}\frac{1}{4x-2}\left(\sqrt{\frac{8x^2+x-3}{x+4}}\right)\)
ai đó giúp với xin cảm ơn nhiều
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Tìm \(\lim\limits_{x->-\infty}\)\(\frac{\left|x\right|\sqrt{4x^2+3}}{2x-1}\)
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lim (n-2-\(\sqrt{3n^2+n-1}\))
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giúp với
tính: lim(x-->1)\(\frac{\sqrt[3]{x}-1}{\sqrt[4]{x}-1}\)
\(\lim\limits_{x\rightarrow3^-}\frac{\sqrt{x^2-7x+12}}{\sqrt{9-x^2}}\)
giúp mình bài này với , mình xin cảm ơn
\(\lim\limits_{x\rightarrow0}\left(\frac{1}{x}+\frac{1}{x^2}\right)\)
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\(\lim\limits_{x\rightarrow-\infty}\frac{|x|+\sqrt{x^2+x}}{x+10}\)