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giải đáp
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bài toán tìm lim, mọi người giúp với nha
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$\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{x^3+1}-\left ( x^2+1 \right )^{\frac{1}{3}}}{x^3}$ $=\mathop {\lim }\limits_{x \to 0}\frac{\sqrt{x^3+1}-1+1-\sqrt[3]{x^2+1}}{x^3}$ $=\mathop {\lim }\limits_{x \to 0}\frac{\frac{x^3}{\sqrt{x^3+1}+1}-\frac{x^2}{1+\sqrt[3]{x^2+1}+\sqrt[3]{(x^2+1)^2}}}{x^3}$ $=\mathop {\lim }\limits_{x \to 0}\frac{\frac{x}{\sqrt{x^3+1}+1}-\frac{1}{1+\sqrt[3]{x^2+1}+\sqrt[3]{(x^2+1)^2}}}{x}$ $=- \infty$
Bạn ktra rồi nếu đúng Click V dumfmk nhé ! Thanks |
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giải đáp
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giúp với ạ
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$=\mathop {\lim }\limits_{x \to 1}\frac{\sqrt{x^3+5x^2+7x+3}-\sqrt{9x^3+7x^2}-(x-1)\sqrt{9x+7}}{x-1}$ $=\mathop {\lim }\limits_{x \to 1}\frac{\frac{(x-1)(-8x^2-10x-3)}{(x+1)\sqrt{x+3}+x\sqrt{9x+7}}-(x-1)\sqrt{9x+7}}{x-1}$ $=\mathop {\lim }\limits_{x \to 1}\left ( \frac{-8x^2-10x-3}{(x+1)\sqrt{x+3}+x\sqrt{9x+7}}-\sqrt{9x+7}\right )$
$=\frac{-53}{8}$
Ktra nếu thấy đúng kq Click V dùm mk nhé ! Thanks |
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giải đáp
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giúp với ạ
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$\mathop {\lim }\limits_{x \to 0}\frac{sinx.cosx-sinx}{sin\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0}\frac{sinx(cosx-1)}{sin\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0}\frac{sinx.\left ( -2sin^2\frac{x}{2} \right )}{sin\frac{x}{2}}$
$=\mathop {\lim }\limits_{x \to 0}\left ( -2.sinx.sin\frac{x}{2}\right )$
$=0$
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giải đáp
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giúp với ạ
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$=\mathop {\lim }\limits_{x \to 0}\frac{x.\left ( \frac{sinx}{x} \right )}{2x.\left ( \frac{tan2x}{2x} \right )}$
$=\frac{1}{2}$
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giải đáp
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giúp với ạ
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$\mathop {\lim }\limits_{x \to 0}\frac{sin5x.sin3x.sinx}{45x^3}$
$=\mathop {\lim }\limits_{x \to 0}\frac{15x^3.\left( \frac{sin5x}{5x} \right )\left ( \frac{sin3x}{3x} \right )\left ( \frac{sinx}{x} \right )}{45x^3}$
$=\frac{1}{3}$
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giải đáp
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giúp với ạ
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$\mathop {\lim }\limits_{x \to 0}\frac{1-\sqrt{cosx}}{1-cos\sqrt{x}}$
$=\mathop {\lim }\limits_{x \to 0}\frac{1-cosx}{(1-cos\sqrt{x})(1+\sqrt{cosx})}$
$=\mathop {\lim }\limits_{x \to 0}\frac{2sin^2\frac{x}{2}}{2sin^2\frac{\sqrt{x}}{2}.(1+\sqrt{cosx})}$
$=\mathop {\lim }\limits_{x \to 0}\frac{\left ( \frac{sin\frac{x}{2}}{\frac{x}{2}}\right )^2.(\frac{x}{2})^2}{\left ( \frac{sin\frac{\sqrt{x}}{2}}{\frac{\sqrt{x}}{2}} \right )^2.\left ( \frac{\sqrt{x}}{2}\right )^2}.\frac{1}{1+\sqrt{cosx}}$
$=0$
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giải đáp
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giúp với ạ
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Tính $\mathop {\lim }\limits_{x \to 1}\frac{x\sqrt[2017]{x}-2008\sqrt[2007]{x}+2007}{(\sqrt[2007]{x}-1)^2}$
Đặt : $t=\sqrt[2007]{x} \Rightarrow $ Khi $x\rightarrow 1$ thì $t\rightarrow 1$
$=\mathop {\lim }\limits_{t \to 1}\frac{t^{2018}-2008t+2007}{(t-1)^2}$
$=\mathop {\lim }\limits_{t \to 1}\frac{t^{2008}-1-2008(t-1)}{(t-1)^2}$$=\mathop {\lim }\limits_{t \to 1}\frac{t^{2007}+t^{2006}+...+t+1-2008}{t-1}$ $=\mathop {\lim }\limits_{t \to 1}\frac{t^{2007}+...+t-2007}{t-1}$ $=\mathop {\lim }\limits_{t \to 1}\frac{t^{2007}-1+t^{2006}-1+...+t-1}{t-1}$ $=\mathop {\lim }\limits_{t \to 1}\left[ {(t^{2006}+t^{2005}+...+t+1})+(t^{2005}+t^{2004}+...+t+1)+...+(t+1)+1 \right]$
$=2007+2006+2005+...+1$
$=1004.2007$
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giải đáp
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$\color{red}{Happy}$ $\color{red}{ New}$ $\color{red}{ Year} $ $\color{red}{2017 ! }$
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$BT=\mathop {\lim }\limits_{x \to 1}\frac{\sqrt{x+3}-2+\sqrt[3]{2x+6}-2}{\sqrt[3]{x-2}+1+\sqrt{3x-2}-1}$
$=\mathop {\lim }\limits_{x \to 1}\frac{\frac{x-1}{\sqrt{x+3}+2}+\frac{2(x-1)}{\sqrt[3]{(2x+6)^2}+\sqrt[3]{2x+6}+4}}{\frac{x-1}{\sqrt[3]{(x-2)^2}-\sqrt[3]{x-2}+1}+\frac{3(x-1)}{\sqrt{3x-2}+1}}$
$=\mathop {\lim }\limits_{x \to 1}\frac{\frac{1}{\sqrt{x+3}+2}+\frac{2}{\sqrt[3]{(2x+6)^2}+2\sqrt[3]{2x+6}+4}}{\frac{1}{\sqrt[3]{(x-2)^2}-\sqrt[3]{x-2}+1}+\frac{3}{\sqrt{3x-2}+1}}=\frac{5}{22}$
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