Lời giải:
\(\lim\limits_{n\to +\infty}(a\sqrt{n+1}+b\sqrt{n+2}+c\sqrt{n+3})=a\lim\limits_{n\to +\infty}\sqrt{n+1}+b\lim\limits_{n\to +\infty}\sqrt{n+2}+c\lim\limits_{n\to +\infty}\sqrt{n+3}\)
\(=a.(+\infty)+b(+\infty)+c.(+\infty)=(a+b+c)(+\infty)\)
Bài 2: Giới hạn của hàm số
Các câu hỏi tương tự
tìm các giới hạn sau
a,\(lim\frac{\left(n^2+1\right)\left(2n+3\right)}{\sqrt{n^4-n^2+1}}\)
b, lim\(\frac{\left(-3^n-6^n\right)}{\left(-3\right)^{n+1}-5^{n+1}}\)
c,lim\(\left(\sqrt{n^4+1}+n-1\right)\)
d, \(\sqrt[3]{1+2n-n^3}\)
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,\(lim\left(\sqrt[3]{n^3+3n^2}-n\right)\)
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 \(\left(\sqrt{n^2-2}-\sqrt[3]{n^3+2n}\right)\)
5 tháng 2 2021 lúc 23:03
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
Tính
1, \(lim\sqrt{n}\left(\sqrt{n+1}-\sqrt{n}\right)\)
2, \(lim\left(\sqrt[3]{n^3+1}-n\right)\)
\(lim\left(\sqrt[3]{n^3+1}\sqrt{n^2+1}-n^2\right)\)
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. \(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}\)
Tính\(\lim\limits_{x\rightarrow+\infty}\left(\sqrt[3]{n^3+1}-\sqrt[3]{n^3+2}\right)\)
6 tháng 2 2021 lúc 14:54
Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)