\(a=\lim\limits_{x\rightarrow3}\dfrac{2x+3-x^2}{\left(x^2-4x+3\right)\left(\sqrt[]{2x+3}+x\right)}=\lim\limits_{x\rightarrow3}\dfrac{\left(x-3\right)\left(-x-1\right)}{\left(x-3\right)\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}\)

\(=\lim\limits_{x\rightarrow3}\dfrac{-x-1}{\left(x-1\right)\left(\sqrt[]{2x+3}+x\right)}=...\)

\(b=\lim\limits_{x\rightarrow0}\dfrac{\left(x+1\right)^{\dfrac{1}{3}}-1}{\left(2x+1\right)^{\dfrac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{3}\left(x+1\right)^{-\dfrac{2}{3}}}{\dfrac{1}{2}\left(2x+1\right)^{-\dfrac{3}{4}}}=\dfrac{2}{3}\)

\(c=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[]{1+4x}-2x-1\right)+\left(2x+1-\sqrt[3]{1+6x}\right)}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{-4x^2}{2x+1+\sqrt[]{4x+1}}+\dfrac{x^2\left(8x+12\right)}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}}{x^2}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{-4}{2x+1+\sqrt[]{4x+1}}+\dfrac{8x+12}{\left(2x+1\right)^2+\left(2x+1\right)\sqrt[3]{1+6x}+\sqrt[3]{\left(1+6x\right)^2}}\right)=...\)

\(a=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-1+1-\sqrt[3]{2x+1}}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{4x}{\sqrt[]{4x+1}+1}+\dfrac{-2x}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{4x+1}+1}+\dfrac{-2}{1+\sqrt[3]{2x+1}+\sqrt[3]{\left(2x+1\right)^2}}\right)=...\)

\(b=\lim\limits_{x\rightarrow1}\dfrac{4\left(x-1\right)\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(x-1\right)\left(\sqrt[]{4x+5}+3\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{4\left(\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4\right)}{5\left(\sqrt[]{4x+5}+3\right)}=...\)

\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(2x+3\right)^{\dfrac{1}{4}}+\left(2+3x\right)^{\dfrac{1}{3}}}{\left(x+2\right)^{\dfrac{1}{2}}-1}\)

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

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

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

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

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

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

\(=\dfrac{3}{1+1+1}+\dfrac{2}{1+1}\)

=1+1

=2

\(\lim\limits_{x\rightarrow0}\dfrac{2x}{\sqrt{4x^2+x^3}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2x}{x\cdot\sqrt{4+x}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{2}{\sqrt{4+x}}=\dfrac{2}{\sqrt{4+0}}=1\)

Không tồn tại do \(\lim\limits_{x\rightarrow0^-}\dfrac{2x}{\sqrt{4x^2+x^3}}\ne\lim\limits_{x\rightarrow0^+}\dfrac{2x}{\sqrt{4x^2+x^3}}\)

\(a=\lim\limits_{x\rightarrow2}\dfrac{\left(x^2-x-2\right)\left(x^2+x\sqrt[3]{3x+2}+\sqrt[3]{\left(3x+2\right)^2}\right)}{\left(x^3-3x-2\right)\left(x+\sqrt[]{x+2}\right)}\)

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

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

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

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

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

\(=...\)

\(c=\lim\limits_{x\rightarrow-1}\dfrac{\left(\sqrt[]{5+4x}-2x-3\right)+\left(2x+3-\sqrt[3]{7+6x}\right)}{x^3+x^2-x-1}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{5+4x-\left(2x+3\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(2x+3\right)^3-\left(7+6x\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4\left(x+1\right)^2}{2x+3+\sqrt[]{5+4x}}+\dfrac{\left(x+1\right)^2\left(8x+20\right)}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{\left(x-1\right)\left(x+1\right)^2}\)

\(=\lim\limits_{x\rightarrow-1}\dfrac{\dfrac{-4}{2x+3+\sqrt[]{5+4x}}+\dfrac{8x+20}{\left(2x+3\right)^2+\left(2x+3\right)\sqrt[3]{7+6x}+\sqrt[3]{\left(7+6x\right)^2}}}{x-1}\)

\(=...\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[]{1+2x}\left(\sqrt[3]{3x+1}-\left(x+1\right)\right)+\left(x+1\right)\left(\sqrt[]{1+2x}-\left(x+1\right)\right)+x^2+\left(2x+1-\sqrt[]{4x+1}\right)}{1+x-\sqrt[]{2x+1}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[]{1+2x}.\dfrac{x^2\left(-x-3\right)}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\dfrac{x^2.\left(x+1\right)}{\sqrt[]{1+2x}+x+1}+x^2+\dfrac{x^2}{2x+1+\sqrt[]{4x+1}}}{\dfrac{x^2}{1+x+\sqrt[]{2x+1}}}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{\left(-x-3\right)\sqrt[]{1+2x}}{\sqrt[3]{\left(3x+1\right)^2}+\sqrt[3]{3x+1}+1}+\dfrac{x+1}{\sqrt[]{1+2x}+x+1}+1+\dfrac{1}{2x+1+\sqrt[]{4x+1}}}{\dfrac{1}{1+x+\sqrt[]{2x+1}}}\)

\(=3\)

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}+\sqrt{5x+4}-5}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt{2x+2}-2+\sqrt{5x+4}-3}{x-1}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{2x+2}+2}+\dfrac{5\left(x-1\right)}{\sqrt{5x+4}+3}}{x-1}=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\sqrt{2x+2}+2}+\dfrac{5}{\sqrt{5x+4}+3}\right)=\dfrac{2}{2+2}+\dfrac{5}{3+3}=...\)

Đề câu b là \(...\sqrt{90-6x}\) hay \(\sqrt{9-6x}\) vậy em? Hình như cái sau mới có lý

a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)

b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)

c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)

d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)

e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)

f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)

g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)

h)

\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)

k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)