1: \(A=\dfrac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)

\(=\dfrac{x^2-xa-x+a}{\left(x-a\right)\left(x^2+ax+a^2\right)}\)

\(=\dfrac{\left(x-a\right)\left(x-1\right)}{\left(x-a\right)\left(x^2+ax+a^2\right)}=\dfrac{x-1}{x^2+ax+a^2}\)

\(lim_{x->a}A=lim_{x->a}\left(\dfrac{x-1}{x^2+ax+a^2}\right)\)

\(=\dfrac{a-1}{a^2+a^2+a^2}=\dfrac{a-1}{3a^2}\)

2: \(B=\dfrac{1}{1-x}-\dfrac{3}{1-x^3}\)

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

\(=\dfrac{-\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{-x-2}{x^2+x+1}\)

\(lim_{x->1}\left(B\right)=\dfrac{-1-2}{1^2+1+1}=\dfrac{-3}{3}=-1\)

3: \(C=\dfrac{\left(x+h\right)^3-x^3}{h}=\dfrac{\left(x+h-x\right)\left(x^2+2xh+h^2+x^2+hx+x^2\right)}{h}\)

\(=3x^2+3hx\)

\(lim_{h->0}\left(C\right)=3x^2+3\cdot0\cdot x=3x^2\)

Câu a.

\(^{lim}_{x\rightarrow3}\dfrac{\sqrt{x+1}-x+1}{x^2-5x+6}\)

Nhân liên hợp ta đc:

\(^{lim}_{x\rightarrow3}\dfrac{x+1-\left(x-1\right)^2}{(x^2-5x+6)\cdot\left(\sqrt{x+1}+x-1\right)}\)

\(=^{lim}_{x\rightarrow3}\dfrac{-x^2+3x}{\left(x-3\right)\left(x-2\right)\left(\sqrt{x+1}+x-1\right)}\)

\(=^{lim}_{x\rightarrow3}\dfrac{-x}{\left(x-2\right)\cdot\left(\sqrt{x+1}+x-1\right)}\)

\(=\dfrac{-3}{\left(3-2\right)\cdot\left(\sqrt{3+1}+3-1\right)}=-\dfrac{3}{4}\)

Câu b.

\(^{lim}_{x\rightarrow-2}\left|x^3-3x\right|\)

\(=\left|\left(-2\right)^3-3\cdot\left(-2\right)\right|=\left|-2\right|=2\)

Câu này đơn giản chỉ thay số thôi nhé, nó ở dạng đa thức nữa!

\(lim_{x\rightarrow1}\frac{x^3+2x-3}{x^2-x}\)   

\(=lim_{x\rightarrow1}\frac{\left(x-1\right)\left(x^2+x+3\right)}{x\left(x-1\right)}\)   

\(=lim_{x\rightarrow1}\frac{x^2+x+3}{x}\)   

\(=\frac{1^2+1+3}{1}\)   

\(=5\)   

\(lim_{x\rightarrow1}\frac{\sqrt{2x+2}-\sqrt{3x+1}}{x-1}\)   

\(=lim_{x\rightarrow1}\frac{\left(2x+2\right)-\left(3x+1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{2x+2-3x-1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-x+1}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1\left(x-1\right)}{\left(x-1\right)\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=lim_{x\rightarrow1}\frac{-1}{\left(\sqrt{2x+2}+\sqrt{3x+1}\right)}\)   

\(=\frac{-1}{\sqrt{2\cdot1+2}+\sqrt{3\cdot1+1}}\)   

\(=\frac{-1}{2+2}=\frac{-1}{4}\)

https://drive.google.com/file/d/14Q-YI3szy-rePnIHWGD35RKCWiCXCT6k/view?usp=sharing

https://drive.google.com/file/d/1425SNt8hu4qt2y1kIcnhIvcxPfODsY1T/view?usp=sharing

làm bài lim xem nào :))) 

P/s Sở Kiều :))

Hướng làm thôi chứ gõ công thức lâu vl

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

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

Đến đây liên hợp là xong phần của bạn đó ;) 

a) \(\mathop {\lim }\limits_{x \to 2} \left( {{x^2} - 4x + 3} \right) = \mathop {\lim }\limits_{x \to 2} {x^2} - \mathop {\lim }\limits_{x \to 2} \left( {4x} \right) + 3 = {2^2} - 4.2 + 3 =  - 1\)

b) \(\mathop {\lim }\limits_{x \to 3} \frac{{{x^2} - 5x + 6}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \frac{{\left( {x - 3} \right)\left( {x - 2} \right)}}{{x - 3}} = \mathop {\lim }\limits_{x \to 3} \left( {x - 2} \right) = \mathop {\lim }\limits_{x \to 3} x - 2 = 3 - 2 = 1\)

c) \(\mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{{\sqrt 1  + 1}} = \frac{1}{2}\)

Lời giải:

a. \(\lim\limits_{x\to 1+}(x^3+x+1)=3>0\)

\(\lim\limits_{x\to 1+}(x-1)=0\) và $x-1>0$ khi $x>1$

\(\Rightarrow \lim\limits_{x\to 1+}\frac{x^3+x+1}{x-1}=+\infty\)

b.

 \(\lim\limits_{x\to -1+}(3x+2)=-1<0\)

\(\lim\limits_{x\to -1+}(x+1)=0\) và $x+1>0$ khi $x>-1$

\(\Rightarrow \lim\limits_{x\to -1+}\frac{3x+2}{x+1}=-\infty\)

c.

\(\lim\limits_{x\to 2-}(x-15)=-17<0\)

\(\lim\limits_{x\to 2-}(x-2)=0\) và $x-2<0$ khi $x<2$

\(\Rightarrow \lim\limits_{x\to 2-}\frac{x-15}{x-2}=+\infty\)

a) Đặt \(f\left( x \right) = 2{x^2} - x\).

Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to 3\) khi \(n \to  + \infty \). Ta có:

\(\lim f\left( {{x_n}} \right) = \lim \left( {2x_n^2 - {x_n}} \right) = 2.\lim x_n^2 - \lim {x_n} = {2.3^2} - 3 = 15\).

Vậy \(\mathop {\lim }\limits_{x \to 3} \left( {2{x^2} - x} \right) = 15\).

b) Đặt \(f\left( x \right) = \frac{{{x^2} + 2x + 1}}{{x + 1}}\).

Hàm số \(y = f\left( x \right)\) xác định trên \(\mathbb{R}\).

Giả sử \(\left( {{x_n}} \right)\) là dãy số bất kì thỏa mãn \({x_n} \to  - 1\) khi \(n \to  + \infty \). Ta có:

\(\lim f\left( {{x_n}} \right) = \lim \frac{{x_n^2 + 2{x_n} + 1}}{{{x_n} + 1}} = \lim \frac{{{{\left( {{x_n} + 1} \right)}^2}}}{{{x_n} + 1}} = \lim \left( {{x_n} + 1} \right) = \lim {x_n} + 1 =  - 1 + 1 = 0\).

Vậy \(\mathop {\lim }\limits_{x \to  - 1} \frac{{{x^2} + 2x + 1}}{{x + 1}} = 0\).

Đề bị lỗi công thức rồi bạn. Bạn cần viết lại để được hỗ trợ tốt hơn.

a/ \(\lim\limits_{x\rightarrow2}\dfrac{2+3}{4+2+4}=\dfrac{5}{10}=\dfrac{1}{2}\)

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