Hoặc là bạn liên hợp \(\left(1+3x\right)^3-1+1-\left(1-4x\right)^4\) rồi rút gọn với mẫu
Nhưng như thế thì dài quá, kiến nghị dùng L'Hopital cho lẹ:
\(\lim\limits_{x\rightarrow0}\frac{\left(3x+1\right)^3-\left(1-4x\right)^4}{x}=\lim\limits_{x\rightarrow0}\frac{3.3.\left(1+3x\right)^2-4.\left(-4\right).\left(1-4x\right)^3}{1}=25\)
Tìm các giới hạn sau:
C=\(\lim\limits_{x\rightarrow0}\frac{\left(3x+1\right)^3-\left(1-4x\right)^4}{x}\)
D=\(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+3x\right)^3-\left(1-4x\right)^4}{x}\)
\(\lim\limits_{x\rightarrow2}\dfrac{2x^2-5x+2}{x^3-3x-2}\)
\(\lim\limits_{x\rightarrow1}\dfrac{x^4-3x+2}{x^3+2x-3}\)
a. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
b. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}\)
tìm các giới hạn sau:
a; \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
b, \(\lim\limits_{x\rightarrow+\infty}\frac{\left(2x-3\right)^2\left(4x+7\right)^3}{\left(3x^3+1\right)\left(10x^2+9\right)}\)
c,\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\) ( bài này k hiểu mk tính kiểu gì 1 cái ra \(+\infty\) một cái ra \(-\infty\))
d, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+4x}.\sqrt{1+6x}-1}{x}\)
e, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
1, \(\lim\limits_{x\rightarrow1}\frac{2x^2-3x+1}{x^3-x^2-x+1}\)
2, \(\lim\limits_{x\rightarrow2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}\)
3, \(\lim\limits_{x\rightarrow0}\frac{1-\sqrt[3]{x-1}}{x}\)
4, \(\lim\limits_{x\rightarrow-\infty}\frac{x^2-5x+1}{x^2-2}\)
5, \(\lim\limits_{x\rightarrow+\infty}\frac{2x^2-4}{x^3+3x^2-9}\)
6, \(\lim\limits_{x\rightarrow2^-}\frac{2x-1}{x-2}\)
7, \(\lim\limits_{x\rightarrow3^+}\frac{8+x-x^2}{x-3}\)
8, \(\lim\limits_{x\rightarrow-\infty}\left(8+4x-x^3\right)\)
9, \(\lim\limits_{x\rightarrow-1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}\)
10, \(\lim\limits_{x\rightarrow-\infty}\frac{\left(2x^2+1\right)^2\left(5x+3\right)}{\left(2x^3-1\right)\left(x+1\right)^2}\)
11, \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{x^2+2x}}{x+3}\)
12, \(\lim\limits_{x\rightarrow1}\frac{\sqrt{5-x^3}-\sqrt[3]{x^2+7}}{x^2-1}\)
13, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x+1}+\sqrt{x+4}-3}{x}\)
14, \(\lim\limits_{x\rightarrow0}\frac{\left(x^2+2020\right)\sqrt{1+3x}-2020}{x}\)
15, \(\lim\limits_{x\rightarrow+\infty}\left(2x-\sqrt{4x^2-3}\right)\)
16, \(\lim\limits_{x\rightarrow a}\frac{x^2-\left(a+1\right)x+a}{x^3-a^3}\)
17, \(\lim\limits_{x\rightarrow1}\frac{x^n-nx+n-1}{\left(x-1\right)^2}\)
18, \(f\left(x\right)=\left\{{}\begin{matrix}\frac{x^2-2x}{8-x^3}\\\frac{x^4-16}{x-2}\end{matrix}\right.\) khi x>2,khi x<2 tại x=2
Bài 1
a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}\)
b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(1-x\right)^2}\)
c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}\)
Bài 2
a. \(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}\)
b. \(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)\)
Tìm \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{\left(2x+1\right)\left(3x+1\right)\left(4x+1\right)}-1}{x}\)
Bài 1
a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}\)
b. \(\lim\limits_{x\rightarrow1}\frac{x^6-5x^5+x}{\left(1-x\right)^2}\)
c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)\left(1+3x\right)-1}{x}\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+x\right)}{x^5+x^2}\)
Bài 2
a. \(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}\)
b. \(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}\left(n\in Z^+,a\ne0\right)\)
c. \(\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)-1}{x}\)
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+3x^2}-\sqrt{x^2-2x}\right)\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
