a) \(\overset{lim}{x\rightarrow1}\dfrac{x^3+2x-3}{x^2-x}=\overset{lim}{x\rightarrow1}\dfrac{\left(x^2+x+3\right)\left(x-1\right)}{x\left(x-1\right)}=\overset{lim}{x\rightarrow1}\dfrac{x^2+x+3}{x}=\dfrac{1^2+1+3}{1}=5\)

b)\(\overset{lim}{x\rightarrow1}\dfrac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}=\overset{lim}{x\rightarrow1}\dfrac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)​ 

                                               = \(\overset{lim}{x\rightarrow1}\dfrac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\) 
                                               = ​\(\overset{lim}{x\rightarrow1}\dfrac{-\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)

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

                                              = \(\dfrac{-1}{\sqrt{2.1+2}+\sqrt{3.1+1}}=-\dfrac{1}{4}\)

\(y'=\left(x.\tan x\right)'=\tan x+x+x.\tan^2x\)

\(y''=\left(\tan x+x+x.\tan^2x\right)'=2+2\tan^2x+2x.\tan x+2x.\tan^3x\)

\(2\left(x^2+y^2\right)\left(1+y\right)=2\left(x^2+x^2.tan^2x\right)\left(1+x.\tan x\right)\)

                               = \(x^2\left(2+2\tan^2x\right)\left(1+x.\tan x\right)\) 

                               = \(x^2\left(2+2x.\tan x+2\tan^2x+2x.\tan^3x\right)\)

                               = x².y''

y_o = 1³ - 3.1² + 1 - 1 = -2

y' = (x³ - 3x² + x - 1)' = 3x² - 6x + 1

y'(x_o) = 3.1² - 6.1 + 1 = -2

Phương trình tiếp tuyến có dạng: y = y'(x_o)(x - x_o) - y_o = -2(x - 1) -2 = -2x

a) \(\overset{lim}{x\rightarrow3}\left(x+2\right)=3+2=5\)

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

Vì\(\left\{{}\begin{matrix}\overset{lim}{x\rightarrow+\infty}\left(x\right)=+\infty\\\overset{lim}{x\rightarrow+\infty}\left(1-\dfrac{1}{x}+\dfrac{1}{x^2}\right)=1>0\end{matrix}\right.\)

y = f(x) = x³ + 1

y_o = 1³ + 3 = 4

y' = (x³ + 3)' = 3x²

y'(x_o) = 3.1² = 3

Phương trình tiếp tuyến có dạng: y = y'(x_o)(x - x_o) + y_o = 3(x - 1)  + 4 = 3x + 1