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a/ \(=lim\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\infty}=0\)
b/ \(=lim\frac{6n+1}{\sqrt{n^2+5n+1}+\sqrt{n^2-n}}=\frac{6+\frac{1}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{1}{n}}}=\frac{6}{1+1}=3\)
c/ \(=lim\frac{6n-9}{\sqrt{3n^2+2n-1}+\sqrt{3n^2-4n+8}}=lim\frac{6-\frac{9}{n}}{\sqrt{3+\frac{2}{n}-\frac{1}{n^2}}+\sqrt{3-\frac{4}{n}+\frac{8}{n^2}}}=\frac{6}{\sqrt{3}+\sqrt{3}}=\sqrt{3}\)
d/ \(=lim\frac{\left(\frac{2}{6}\right)^n+1-4\left(\frac{4}{6}\right)^n}{\left(\frac{3}{6}\right)^n+6}=\frac{1}{6}\)
e/ \(=lim\frac{\left(\frac{3}{5}\right)^n-\left(\frac{4}{5}\right)^n+1}{\left(\frac{3}{5}\right)^n+\left(\frac{4}{5}\right)^n-1}=\frac{1}{-1}=-1\)
f/ Ta có công thức:
\(1+3+...+\left(2n+1\right)^2=\left(n+1\right)^2\)
\(\Rightarrow lim\frac{1+3+...+2n+1}{3n^2+4}=lim\frac{\left(n+1\right)^2}{3n^2+4}=lim\frac{\left(1+\frac{1}{n}\right)^2}{3+\frac{4}{n^2}}=\frac{1}{3}\)
g/ \(=lim\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n}-\frac{1}{n+1}\right)=lim\left(1-\frac{1}{n+1}\right)=1-0=1\)
h/ Ta có: \(1^2+2^2+...+n^2=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
\(\Rightarrow lim\frac{n\left(n+1\right)\left(2n+1\right)}{6n\left(n+1\right)\left(n+2\right)}=lim\frac{2n+1}{6n+12}=lim\frac{2+\frac{1}{n}}{6+\frac{12}{n}}=\frac{2}{6}=\frac{1}{3}\)
a/ \(=lim\frac{\left(-\frac{2}{3}\right)^n+1}{-2.\left(-\frac{2}{3}\right)^n+3}=\frac{1}{3}\)
b/ \(=lim\frac{\left(2-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\left(3+\frac{4}{n}\right)}{\left(\frac{5}{n}-6\right)^3}=\frac{2.1.3}{\left(-6\right)^3}=-\frac{1}{36}\)
c/ \(=lim\frac{5n+3}{\sqrt{n^2+5n+1}+\sqrt{n^2-2}}=\frac{5+\frac{3}{n}}{\sqrt{1+\frac{5}{n}+\frac{1}{n^2}}+\sqrt{1-\frac{2}{n}}}=\frac{5}{1+1}=\frac{5}{2}\)
d/ \(=lim\frac{5.\left(\frac{1}{2}\right)^n-6}{4.\left(\frac{1}{3}\right)^n+1}=\frac{-6}{1}=-6\)
e/ \(=-n^3\left(2+\frac{3}{n}-\frac{5}{n^2}+\frac{2020}{n^3}\right)=-\infty.2=-\infty\)
\(A=lim\frac{\sqrt{n+2}+\sqrt{n+1}}{1}=lim\left[n\left(\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{1}{n}}\right)\right]=+\infty.2=+\infty\)
\(B=lim\frac{8^3.64^n-9.27^n}{4^4.64^n+5^3.25^n}=\frac{8^3-9.\left(\frac{27}{64}\right)^n}{4^4+5^3\left(\frac{25}{64}\right)^n}=\frac{8^3}{4^4}=2\)
\(1;-\frac{1}{2};\frac{1}{4}...\) là dãy cấp số nhân lùi vô hạn có \(u_1=1\) và \(q=-\frac{1}{2}\)
Do \(\left|q\right|< 1\) nên theo công thức tổng cấp số nhân:
\(S_n=\frac{u_1}{1-q}=\frac{1}{1+\frac{1}{2}}=\frac{2}{3}\)
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?
\(=lim\frac{2.2^{5n}+3}{9.3^{5n}+1}=lim\frac{2.\left(\frac{2}{3}\right)^{5n}+3\left(\frac{1}{3}\right)^{5n}}{9+\left(\frac{1}{3}\right)^{5n}}=\frac{0}{9}=0\)
\(b=lim\frac{\left(-\frac{1}{3}\right)^n+4}{-1\left(-\frac{1}{3}\right)^n-2}=\frac{4}{-2}=-2\)
\(c=1+lim\frac{-n}{n^2+\sqrt{n^4+n}}=1+lim\frac{-\frac{1}{n}}{1+\sqrt{1+\frac{1}{n^3}}}=1+\frac{0}{2}=1\)
\(-2\le2cosn^2\le2\Rightarrow\frac{-2}{n^2+1}\le\frac{2cosn^2}{n^2+1}\le\frac{2}{n^2+1}\)
Mà \(lim\frac{-2}{n^2+1}=lim\frac{2}{n^2+1}=0\Rightarrow lim\frac{2cosn^2}{n^2+1}=0\)
\(d=lim\left[n\left(\sqrt{1-\frac{2}{n^2}}-1+1-\sqrt[3]{1+\frac{2}{n^2}}\right)\right]\)
\(=lim\left[n\left(\frac{-\frac{2}{n^2}}{\sqrt{1-\frac{2}{n^2}}+1}-\frac{\frac{2}{n^2}}{\sqrt[3]{\left(1+\frac{2}{n^2}\right)^2}+\sqrt[3]{1+\frac{2}{n^2}}+1}\right)\right]\)
\(=lim\left(\frac{-\frac{2}{n}}{\sqrt{1-\frac{2}{n^2}}+1}-\frac{\frac{2}{n}}{\sqrt[3]{\left(1+\frac{2}{n^2}\right)^2}+\sqrt[3]{1+\frac{2}{n^2}}+1}\right)=\frac{0}{2}-\frac{0}{1+1+1}=0\)
a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)
b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))
= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )
= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)
= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)
= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)
= lim \(-3n=-\infty\)
c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
\(a=lim\frac{n^2+n}{6n^3}=lim\frac{\frac{1}{n}+\frac{1}{n^3}}{6}=\frac{0}{6}=0\)
\(b=lim\frac{1+\frac{2}{n}}{1+\frac{1}{n}}+lim\frac{sinn}{2^n}=1+0=1\)
Giải thích: \(-1\le sin\left(n\right)\le1\) \(\forall n\Rightarrow\frac{-1}{2^n}\le\frac{sin\left(n\right)}{2^n}\le\frac{1}{2^n}\)
Mà \(lim\frac{-1}{2^n}=lim\frac{1}{2^n}=0\Rightarrow lim\frac{sin\left(n\right)}{2^n}=0\) theo nguyên tắc giới hạn kẹp
\(c=lim\frac{-3n-1}{\sqrt{n^2-3n}+\sqrt{n^2+1}}=lim\frac{-3-\frac{1}{n}}{\sqrt{1-\frac{3}{n}}+\sqrt{1+\frac{1}{n^2}}}=\frac{-3}{1+1}=-\frac{3}{2}\)
\(d=lim\frac{3n^2}{\sqrt[3]{\left(n^3+3n^2\right)^2}+n\sqrt[3]{n^3+3n^2}+n^2}=lim\frac{3}{\sqrt[3]{\left(1+\frac{3}{n}\right)^2}+\sqrt[3]{1+\frac{3}{n}}+1}=\frac{3}{1+1+1}=1\)
a/ \(lim\left(\sqrt[3]{n-n^3}+n+\sqrt{n^2+3n}-n\right)\)
\(=lim\left(\frac{n}{\sqrt[3]{\left(n-n^3\right)^2}-n\sqrt[3]{\left(n-n^3\right)}+n^2}+\frac{3n}{\sqrt{n^2+3n}+n}\right)\)
\(=lim\left(\frac{1}{\sqrt[3]{n^3+2n+\frac{1}{n}}+\sqrt[3]{n^3-n}+n}+\frac{3}{\sqrt{1+\frac{3}{n}}+1}\right)=0+\frac{3}{1+1}=\frac{3}{2}\)
b/ \(lim\left(\frac{-2\sqrt{n}-4}{\sqrt{n-2\sqrt{n}}+\sqrt{n+4}}\right)=lim\left(\frac{-2-\frac{4}{\sqrt{n}}}{\sqrt{1-\frac{2}{\sqrt{n}}}+\sqrt{1+\frac{4}{n}}}\right)=-\frac{2}{1+1}=-1\)
c/ \(lim\left(\frac{3n^2}{\sqrt[3]{n^6+6n^5+9n^4}+\sqrt[3]{n^6+3n^5}+n^2}\right)=lim\left(\frac{3}{\sqrt[3]{1+\frac{6}{n}+\frac{9}{n^2}}+\sqrt[3]{1+\frac{3}{n}}+1}\right)=\frac{3}{3}=1\)
d/ \(lim\left(\sqrt[3]{n^3+6n}-n+n-\sqrt{n^2-4n}\right)=lim\left(\frac{6n}{\sqrt[3]{n^6+12n^4+36n^2}+\sqrt[3]{n^6+6n^4}+n^2}+\frac{4n}{n+\sqrt{n^2-4n}}\right)\)
\(=lim\left(\frac{6}{\sqrt[3]{n^3+12n+\frac{36}{n}}+\sqrt[3]{n^3+6n}+n}+\frac{4}{1+\sqrt{1-\frac{4}{n}}}\right)=0+\frac{4}{1+1}=2\)
e/ \(lim\left(\frac{-3.3^n+4.4^n}{5.3^n+\frac{3}{2}.4^n}\right)=lim\left(\frac{-3\left(\frac{3}{4}\right)^n+4}{5.\left(\frac{3}{4}\right)^n+\frac{3}{2}}\right)=\frac{0+4}{0+\frac{3}{2}}=\frac{8}{3}\)
f/ \(lim\left(\frac{9^n-5.5^n+7.7^n}{9.3^n+5^n+2.8^n}\right)=lim\left(\frac{1-5.\left(\frac{5}{9}\right)^n+7\left(\frac{7}{9}\right)^n}{9.\left(\frac{1}{3}\right)^n+\left(\frac{5}{9}\right)^n+2.\left(\frac{8}{9}\right)^n}\right)=\frac{1}{0}=+\infty\)
g/ \(lim\left(\frac{6.6^n+3^5.9^n}{3^3.9^n-\frac{1}{2}.4^n}\right)=lim\left(\frac{6\left(\frac{2}{3}\right)^n+3^5}{3^3-\frac{1}{2}\left(\frac{4}{9}\right)^n}\right)=\frac{3^5}{3^3}=9\)
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)