# Bài 2: Giới hạn của hàm số

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$\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)$

$\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)$

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

$\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)$

$\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}$

7 tháng 11 2023 lúc 19:46

Em là tám lại ạ

Em là duy khôi ạ

Em là văn tam ạ

Em là mạnh Tuấn ạ

8 tháng 11 2023 lúc 19:00

a: $\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x-2}-\dfrac{12}{x^3-8}\right)$

$=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x+4-12}{\left(x-2\right)\left(x^2+2x+4\right)}$

$=\lim\limits_{x\rightarrow2}\dfrac{x^2+2x-8}{\left(x-2\right)\left(x^2+2x+4\right)}$

$=\lim\limits_{x\rightarrow2}\dfrac{x+4}{x^2+2x+4}$

$=\dfrac{2+4}{2^2+2\cdot2+4}=\dfrac{6}{4+4+4}=\dfrac{6}{12}=\dfrac{1}{2}$

b: $\lim\limits_{x\rightarrow2}\left(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}\right)$

$=\lim\limits_{x\rightarrow2}\left(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}\right)$

$=\lim\limits_{x\rightarrow2}\left(\dfrac{x-3+x-1}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}\right)$

$=\lim\limits_{x\rightarrow2}\dfrac{2x-4}{\left(x-2\right)\left(x-3\right)\left(x-1\right)}$

$=\lim\limits_{x\rightarrow2}\dfrac{2}{\left(x-3\right)\left(x-1\right)}=\dfrac{2}{\left(2-3\right)\left(2-1\right)}=-2$

d: $\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-\sqrt[3]{x^3-1}\right)$

$=\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x^2+1}-x+x-\sqrt[3]{x^3-1}\right)$

$=\lim\limits_{x\rightarrow+\infty}\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}+\dfrac{x^3-x^3+1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}$

$=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{1}{\sqrt{x^2+1}+x}+\dfrac{1}{\sqrt[3]{x^2}+x\cdot\sqrt[3]{x^3-1}+\sqrt[3]{\left(x^3-1\right)^2}}\right)$

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

=0

c: $\lim\limits_{x\rightarrow+\infty}\left[x\cdot\left(\sqrt{x^2+1}-x\right)\right]$

$=\lim\limits_{x\rightarrow+\infty}\left[x\cdot\dfrac{x^2+1-x^2}{\sqrt{x^2+1}+x}\right]$

$=\lim\limits_{x\rightarrow+\infty}\dfrac{x}{\sqrt{x^2+1}+x}$

$=\lim\limits_{x\rightarrow+\infty}\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}+1}=\dfrac{1}{1+1}=\dfrac{1}{2}$

e: $\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}$

$=\lim\limits_{x\rightarrow0}\dfrac{x^2+1-1}{\sqrt{x^2+1}+1}:\dfrac{x^2+16-16}{\sqrt{x^2+16}+4}$

$=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x^2+16}+4}{\sqrt{x^2+1}+1}=\dfrac{4+4}{1+1}=\dfrac{8}{2}=4$

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