Bạn muốn tìm giới hạn nhưng lại không chỉ rõ $n$ chạy đến đâu?
Điển hình như câu 1:
$n\to 0$ thì giới hạn là $3$
$n\to \pm \infty$ thì giới hạn là $\pm \infty$
Bạn phải ghi rõ đề ra chứ?
Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
tìm các giới hạn sau:
a, lim\(\frac{2^{5n+1}+3}{3^{5n+2}+1}\)
b, lim\(\frac{\left(-1\right)^n+4.3^n}{\left(-1\right)^{n+1}-2.3^n}\)
c, lim \(\left(1+n^2-\sqrt{n^4+n}\right)\)
d, lim \(\frac{2cosn^2}{n^2+1}\)
e, lim \(\left(\sqrt{n^2-2}-\sqrt[3]{n^3+2n}\right)\)
tìm các giới hạn sau:
a; lim\(\frac{1+2+3+...+n}{3n^3}\)
b, lim \(\left(\frac{n+2}{n+1}+\frac{sin\text{n}}{2^n}\right)\)
c, lim \(\left(\sqrt{n^2-3n}-\sqrt{n^2+1}\right)\)
d,\(lim\left(\sqrt[3]{n^3+3n^2}-n\right)\)
H24
5 tháng 2 2021 lúc 23:03
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
a. limlimits_{xrightarrow a}frac{xsqrt{x}-asqrt{a}}{sqrt{x}-sqrt{a}} e. limlimits_{xrightarrow0}frac{sqrt{1+x}-sqrt[3]{1+x}}{x}
b. limlimits_{xrightarrow1}frac{sqrt[n]{x}-1}{sqrt[m]{x}-1}left(m,nin Z^+right) f. limlimits_{xrightarrow2}frac{sqrt[3]{8x+11}-sqrt{x+7}}{x^2-3x+2}
c. limlimits_{xrightarrow1}frac{left(1-sqrt{x}right)left(1-sqrt[3]{x}right)left(1-sqrt[4]{x}right)left(1-sqrt[5]{x}right)}{left(1-xright)^4}...
Đọc tiếp
a. \(\lim\limits_{x\rightarrow a}\frac{x\sqrt{x}-a\sqrt{a}}{\sqrt{x}-\sqrt{a}}\) e. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-\sqrt[3]{1+x}}{x}\)
b. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[n]{x}-1}{\sqrt[m]{x}-1}\left(m,n\in Z^+\right)\) f. \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-\sqrt{x+7}}{x^2-3x+2}\)
c. \(\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)\left(1-\sqrt[4]{x}\right)\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)^4}\) g. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-\sqrt{2x-1}}{x^3-1}\)
d. \(\lim\limits_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) h. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}\)
H24
6 tháng 2 2021 lúc 14:54
Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)
Tính
1, \(lim\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\)
2, \(lim\left(\sqrt[3]{n^3+1}-n\right)\)
1) lim \(\frac{3n^2+5n+4}{2-n^2}\)
2) lim \(\frac{2n^3-4n^2+3n+7}{n^3-7n+5}\)
3) lim \(\left(\frac{2n^3}{2n^2+3}+\frac{1-5n^2}{5n+1}\right)\)
4) lim \(\frac{1+3^n}{4+3^n}\)
5) lim \(\frac{4.3^n+7^{n+1}}{2.5^n+7^n}\)
\(lim\left(\sqrt[3]{n^3+1}\sqrt{n^2+1}-n^2\right)\)
Bài 1
a. limlimits_{xrightarrow-1}frac{x^5+1}{x^3+1}
b. limlimits_{xrightarrow1}frac{x^6-5x^5+x}{left(1-xright)^2}
c. limlimits_{xrightarrow0}frac{left(1+xright)left(1+2xright)left(1+3xright)-1}{x}
d. limlimits_{xrightarrow0}frac{left(1+xright)^5-left(1+xright)}{x^5+x^2}
Bài 2
a. limlimits_{xrightarrow1}frac{x^m-1}{x^n-1}
b. limlimits_{xrightarrow a}frac{x-a}{x^n-a^n}left(nin Z^+,ane0right)
c. limlimits_{xrightarrow0}frac{x+x^2+...+x^n-n}{x-1}
d. limlimits_{xrightarrow0}frac{lef...
Đọc tiếp
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. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)-1}{x}\)