Tìm lim √4n+5 -2√n
Giúp mình đc ko mình cần gấp ngày mai nộp rồi T^T thank
Lim căn 9n^+2n+n-2/căn 4n^+1
Lim n/căn 4n^+2+căn n^
Lim căn 4n+2- căn 2n-5/căn n+3
Lim căn 4n^+n+1-n/n^+2
Lim căn 9n^+n+1-2n/3n^+2
\(lim\frac{\sqrt{9n^2+2n}+n-2}{\sqrt{4n^2+1}}=lim\frac{\sqrt{9+\frac{2}{n}}+1-\frac{2}{n}}{\sqrt{4+\frac{1}{n^2}}}=\frac{\sqrt{9}+1}{\sqrt{4}}=2\)
\(lim\frac{n}{\sqrt{4n^2+2}+\sqrt{n^2}}=lim\frac{1}{\sqrt{4+\frac{2}{n^2}}+\sqrt{1}}=\frac{1}{\sqrt{4}+\sqrt{1}}=\frac{1}{3}\)
\(lim\frac{\sqrt{4n+2}-\sqrt{2n-5}}{\sqrt{n+3}}=lim\frac{\sqrt{4+\frac{2}{n}}-\sqrt{2-\frac{5}{n}}}{\sqrt{1+\frac{3}{n}}}=\frac{2-\sqrt{2}}{1}=2-\sqrt{2}\)
l\\(lim\frac{\sqrt{4n^2+n+1}-n}{n^2+2}=lim\frac{\sqrt{4+\frac{1}{n}+\frac{1}{n^2}}-1}{n+\frac{2}{n}}=\frac{1}{\infty}=0\)
\(lim\frac{\sqrt{9n^2+n+1}-2n}{3n^2+2}=\frac{\sqrt{9+\frac{1}{n}+\frac{1}{n^2}}-2}{3n+\frac{2}{n}}=\frac{1}{\infty}=0\)
Muốn giúp bạn lắm mà ko sao dịch được đề :D
Bạn sử dụng công cụ gõ công thức, nó ở ngoài cùng bên trái khung soạn thảo, chỗ khoanh đỏ ấy, cực dễ sử dụng
Tìm giới hạn dãy số sau
\(lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}\)
\(lim\left(3.2^{n+1}-5.3^n+7n\right)\)
\(\lim\dfrac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}=\lim\dfrac{\left(\dfrac{2n-1}{n}\right)\left(\dfrac{3n^2+2}{n^2}\right)^3}{\dfrac{-2n^5+4n^3-1}{n^7}}\)
\(=\lim\dfrac{\left(2-\dfrac{1}{n}\right)\left(3+\dfrac{2}{n^2}\right)^3}{-\dfrac{2}{n^2}+\dfrac{4}{n^4}-\dfrac{1}{n^7}}=-\infty\)
\(\lim3^n\left(6.\left(\dfrac{2}{3}\right)^n-5+\dfrac{7n}{3^n}\right)=+\infty.\left(-5\right)=-\infty\)
Tìm \(lim\left(\sqrt{4n^2+n}-\sqrt{4n^2+2}\right)\)
\(lim\left(\sqrt{4n^2+n}-\sqrt{4n^2+2}\right)\)
\(=lim\dfrac{\left(\sqrt{4n^2+n}-\sqrt{4n^2+2}\right)\times\left(\sqrt{4n^2+n}+\sqrt{4n^2+2}\right)}{\left(\sqrt{4n^2+n}+\sqrt{4n^2+2}\right)}\)
\(=lim\dfrac{\left(\sqrt{4n^2+n}\right)^2-\left(\sqrt{4n^2+2}\right)^2}{\sqrt{4n^2+n}+\sqrt{4n^2+2}}\)
\(=lim\dfrac{4n^2+n-4n^2-2}{\sqrt{4n^2+n}+\sqrt{4n^2+2}}\)
\(=lim\dfrac{n-2}{\sqrt{4n^2+n}+\sqrt{4n^2+2}}\)
\(=lim\dfrac{\dfrac{n}{n}-\dfrac{2}{n}}{\dfrac{n}{n}\sqrt{\dfrac{4n^2}{n^2}+\dfrac{n}{n^2}}+\dfrac{n}{n}\sqrt{\dfrac{4n^2}{n^2}+\dfrac{2}{n^2}}}\)
\(=\dfrac{1-0}{1\sqrt{4+0}+1\sqrt{4+0}}\)
\(=\dfrac{1}{2+2}\)
\(=\dfrac{1}{4}\)
Tìm các giới hạn sau:
a) \(\lim \left( {2 + {{\left( {\frac{2}{3}} \right)}^n}} \right)\);
b) \(\lim \left( {\frac{{1 - 4n}}{n}} \right)\).
a) Đặt \({u_n} = 2 + {\left( {\frac{2}{3}} \right)^n} \Leftrightarrow {u_n} - 2 = {\left( {\frac{2}{3}} \right)^n}\).
Suy ra \(\lim \left( {{u_n} - 2} \right) = \lim {\left( {\frac{2}{3}} \right)^n} = 0\)
Theo định nghĩa, ta có \(\lim {u_n} = 2\). Vậy \(\lim \left( {2 + {{\left( {\frac{2}{3}} \right)}^n}} \right) = 2\)
b) Đặt \({u_n} = \frac{{1 - 4n}}{n} = \frac{1}{n} - 4 \Leftrightarrow {u_n} - \left( { - 4} \right) = \frac{1}{n}\).
Suy ra \(\lim \left( {{u_n} - \left( { - 4} \right)} \right) = \lim \frac{1}{n} = 0\).
Theo định nghĩa, ta có \(\lim {u_n} = - 4\). Vậy \(\lim \left( {\frac{{1 - 4n}}{n}} \right) = - 4\)
Tìm các giới hạn sau:
a) \(lim\left(\sqrt{4n+1}-2\sqrt{n}\right)\)
b) \(lim\left(\sqrt{n^2+2n}-\sqrt{n^2-2n}-n\right)\)
c) \(lim\left(\sqrt{9^n-3^n}-4^n\right)\)
d) \(lim\left(3n^3+2n^2+n\right)\)
\(a=\lim\dfrac{1}{\sqrt{4n+1}+2\sqrt{n}}=\dfrac{1}{\infty}=0\)
\(b=\lim n\left(\sqrt{1+\dfrac{2}{n}}-\sqrt{1-\dfrac{2}{n}}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(c=\lim4^n\left(\sqrt{\left(\dfrac{9}{16}\right)^n-\left(\dfrac{3}{16}\right)^n}-1\right)=+\infty.\left(-1\right)=-\infty\)
\(d=\lim n^3\left(3+\dfrac{2}{n}+\dfrac{1}{n^2}\right)=+\infty.3=+\infty\)
biết \(lim\dfrac{\sqrt{\left(3-4n\right)^2+1}+an-1}{\sqrt{n^2+4n+1}+an}=2\). tìm a
\(\lim\dfrac{\sqrt{\left(3-4n\right)^2+1}+an-1}{\sqrt{n^2+4n+1}+an}=\lim\dfrac{\sqrt{\left(\dfrac{3}{n}-4\right)^2+\dfrac{1}{n}}+a-\dfrac{1}{n}}{\sqrt{1+\dfrac{4}{n}+\dfrac{1}{n^2}}+an}\)
\(=\dfrac{4+a}{1+a}=2\Leftrightarrow4+a=2a+2\Rightarrow a=2\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4x^4-3n^2+4\right)\)
1) \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}=\lim\limits_{n\rightarrow\infty}\dfrac{2n\left(1-\dfrac{4}{n}\right)}{n\left(1-\dfrac{1}{n}\right)}=2\)
2) \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}=\dfrac{1}{4n}=\infty\)
3) \(\lim\limits_{n\rightarrow\infty}\left(-2n^5+4n^4-3n^2+4\right)=\lim\limits_{n\rightarrow\infty}n^5\left(-2+\dfrac{4}{n}-\dfrac{3}{n^2}+\dfrac{4}{n^5}\right)=-2n^5=-\infty\)
1) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
2) tính \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
1: \(\lim\limits_{n\rightarrow\infty}\dfrac{6n-8}{n-1}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{6-\dfrac{8}{n}}{1-\dfrac{1}{n}}=\dfrac{6-0}{1-0}\)
\(=\dfrac{6}{1}=6\)
2: \(\lim\limits_{n\rightarrow\infty}\dfrac{n^2+5n-3}{4n^3-2n+5}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{n^2\left(1+\dfrac{5}{n}-\dfrac{3}{n^2}\right)}{n^3\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\)
\(=\lim\limits_{n\rightarrow\infty}\left(\dfrac{1}{n}\cdot\dfrac{1+\dfrac{5}{n}-\dfrac{3}{n^2}}{\left(4-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}\right)\)
=0
Tìm a,b để:
1, Lim(\(\sqrt{4n^2+2n+1}\)-an+b)=1
2, Lim( \(\sqrt{n^2+6n-1}-\sqrt{an^2+bn+2}\))=4
1.
\(\lim\left(\sqrt{4n^2+2n+1}-\left(an-b\right)\right)=\lim\dfrac{4n^2+2n+1-\left(an-b\right)^2}{\sqrt{4n^2+2n+1}+an-b}\)
\(=\lim\dfrac{\left(4-a^2\right)n^2+\left(2+ab\right)n+1-b^2}{\sqrt{4n^2+2n+1}+an-b}\)
\(=\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}\)
- Nếu \(4-a^2\ne0\Rightarrow\) giới hạn đã cho đạt giá trị dương vô cực \(\Rightarrow\) ktm
\(\Rightarrow4-a^2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\end{matrix}\right.\)
- Với \(a=-2\Rightarrow\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}=-\infty\) (ktm)
- Với \(a=2\Rightarrow\lim\dfrac{\left(4-a^2\right)n+2+ab+\dfrac{1-b^2}{n}}{\sqrt{4+\dfrac{2}{n}+\dfrac{1}{n^2}}+a-\dfrac{b}{n}}=\dfrac{2+2b}{4}\)
\(\Rightarrow\dfrac{b+1}{2}=1\Rightarrow b=1\)
Vậy \(a=2;b=1\)
Câu 2 làm tương tự
đặt \(a=lim\dfrac{3n^3-2n+1}{4n^4+2n+1}\). tìm \(lim\dfrac{an^3-\left(a+2\right)n^2+1}{4an^3-n^2+3n+3}\)
\(a=\lim\limits\dfrac{3n^3-2n+1}{4n^4+2n+1}=\lim\limits\dfrac{\dfrac{3n^3}{n^4}-\dfrac{2n}{n^4}+\dfrac{1}{n^4}}{\dfrac{4n^4}{n^4}+\dfrac{2n}{n^4}+\dfrac{1}{n^4}}=0\)
\(\Rightarrow\lim\limits\dfrac{-2n^2+1}{-n^2+3n+3}=\lim\limits\dfrac{-\dfrac{2n^2}{n^2}+\dfrac{1}{n^2}}{-\dfrac{n^2}{n^2}+\dfrac{3n}{n^2}+\dfrac{3}{n^2}}=-\dfrac{2}{-1}=2\)