Tuyển Cộng tác viên Hoc24 nhiệm kì 26 tại đây: https://forms.gle/dK3zGK3LHFrgvTkJ6
Lời giải:
\(\lim\limits _{x\to +\infty}\sqrt{\frac{3x^4+4x^5+2}{9x^5+5x^4+4}}=\lim\limits _{x\to +\infty}\sqrt{\frac{\frac{3}{x}+4+\frac{2}{x^5}}{9+\frac{5}{x}+\frac{4}{x^5}}}=\sqrt{\frac{4}{9}}=\frac{2}{3}\)
Đáp án B.
Bài 1
a. \(\lim\limits_{x\rightarrow-\infty}\frac{\sqrt{4x^2}+1}{3x-1}\)
b. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9x^2+x+1}-\sqrt{4x^2+2x+1}}{x+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+2x+3}+4x+1}{\sqrt{4x^2+1}+2-x}\)
d. \(\lim\limits_{x\rightarrow+\infty}\frac{3x-2\sqrt{x}+\sqrt{x^4-5x}}{2x^2+4x-5}\)
Bài 2
a. \(\lim\limits_{x\rightarrow-\infty}\frac{2x+1}{x-1}\)
b. \(\lim\limits_{x\rightarrow-\infty}\frac{2x^3+3}{x^3-2x^2+1}\)
c. \(\lim\limits_{x\rightarrow+\infty}\frac{\left(3x^2+1\right)\left(5x+3\right)}{\left(2x^3-1\right)\left(x+4\right)}\)
Tính các giới hạn sau đây :
\(L_1=lim\frac{x^3+3x^2-2x}{x^5+4x}\left(x\rightarrow0\right)\)
\(L_2=lim\frac{x^3-3x+2}{\left(4-2x\right)^3}\left(x\rightarrow+\infty\right)\)
\(L_3=lim\frac{2x^2+3x+1}{x^2+x}\left(x\rightarrow-1\right)\)
\(L_4=lim\frac{x^2-4x+1}{4-x^2}\left(x\rightarrow2\right)\)
\(L_5=lim\frac{\sqrt{x+1}-2}{x-2}\left(x\rightarrow3\right)\)
\(L_6=lim\frac{\sqrt{x+3}-x-1}{x^2-1}\left(x\rightarrow1\right)\)
\(L_7=lim\left(\sqrt{x^2+x+1}-x+1\right)\left(x\rightarrow+\infty\right)\)
\(L_8=lim\left(\sqrt{x^2+x+1}-3x+2\right)\left(x\rightarrow-\infty\right)\)
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 3. Tính các giới hạn sau:
a) \(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^3-5x^2+1}{7x^2-x+4}\)
b) \(\lim\limits_{x\rightarrow+\infty}x\sqrt{\dfrac{x^2+2x+3}{3x^4+4x^2-5}}\)
giới hạn \(lim\frac{\sqrt{x^2+1}+x}{3x+5}\) \(\left(x\rightarrow+\infty\right)\) bằng :
A. \(\frac{2}{3}\)
B. \(\frac{1}{3}\)
C. 0
D. \(+\infty\)
Tùy theo giá trị của tham số m, tính giới hạn: \(\lim\limits_{n\rightarrow-\infty}\left(\sqrt[3]{8x^3+5x^2+1}-\sqrt{9x^2+3x+5}+mx\right)\)
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, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}.\sqrt[3]{1+4x}-1}{x}\)
a/ \(^{lim}_{x->0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b/\(^{lim}_{x->1}\left(\frac{1}{1-x}-\frac{1}{1-x^3}\right)\)
c/ \(^{lim}_{x->+\infty}\left(\sqrt[3]{2x-1}-\sqrt[3]{2x+1}\right)\)
d/ \(^{lim}_{x->-\infty}\left(\sqrt[3]{3x^3-1}+\sqrt{x^2+2}\right)\)
e/\(^{lim}_{x->2}\left(\frac{1}{x^2-3x+2}+\frac{1}{x^2-5x+6}\right)\)
f/ \(^{lim}_{x->0^{+-}}\left(\frac{2x}{\sqrt{4x^2+x^3}}\right)\)
