Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x^2+2}-\sqrt{4+x}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{3x^2-x-2}{\sqrt{3x^2+2}+\sqrt{4+x}}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{3x+2}{\left(x+1\right)\left(\sqrt{3x^2+2}+\sqrt{4+x}\right)}=\dfrac{5}{2.2\sqrt{5}}=\dfrac{\sqrt{5}}{4}\).
Từ đó a = 5; b = 4 nên a - b = 1.
Bạn không nói rõ $n$ tiến tới đâu thì sẽ có nhiều kết quả. Căn cứ vào đáp án mình đoán $n\to +\infty$
Lời giải:
\(\lim\limits_{n\to +\infty}\frac{7\sqrt{3n^2+n}}{2(3n+2)}=\lim\limits_{n\to +\infty}\frac{\frac{7\sqrt{3n^2+n}}{n}}{\frac{2(3n+2)}{n}}=\lim\limits_{n\to +\infty}=\frac{7\sqrt{3+\frac{1}{n}}}{6+\frac{4}{n}}\)\(=\frac{7\sqrt{3}}{6}\)
$\frac{a}{b}$ tối giản $\Rightarrow a=7; b=6$
$\Rightarrow a+b=7+6=13$
Đáp án A.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{5x-1}-2+2-\sqrt[3]{10x-2}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{5\left(x-1\right)}{\sqrt{5x-1}+2}-\frac{10\left(x-1\right)}{4+2\sqrt[3]{10x-2}+\sqrt[3]{\left(10x-2\right)^2}}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{5}{\sqrt{5x-1}+2}-\frac{10}{4+2\sqrt[3]{10x-2}+\sqrt[3]{\left(10x-2\right)^2}}\right)=\frac{5}{12\sqrt{2}}=\frac{5}{3.2^2\sqrt{2}}\)
\(\Rightarrow\left\{{}\begin{matrix}a=5\\b=2\end{matrix}\right.\) \(\Rightarrow a+b=7\)
\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)
Để hàm liên tục tại \(x=1\)
\(\Rightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow k=2\sqrt{2019}\)
2.
\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{x^2-1}=\frac{1}{2}\Leftrightarrow\left\{{}\begin{matrix}a+b+1=0\\\lim\limits_{x\rightarrow1}\frac{2x+a}{2x}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=-1\\\frac{a+2}{2}=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=0\end{matrix}\right.\) \(\Rightarrow S=1\)
3.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)
\(\Rightarrow a+b+c=1+12+0=13\)
a)lim \(\frac{\sqrt{n^2-4n}-\sqrt{4n+1}}{\sqrt{3n^2+1}+n}\)
=lim \(\frac{\sqrt{1-\frac{4}{n}}-\sqrt{\frac{4}{n}+\frac{1}{n^2}}}{\sqrt{3+\frac{1}{n^2}}+1}=\frac{1}{\sqrt{3}+1}\)
b)lim \(\frac{\sqrt[3]{8n^3+n^2}-n}{2n-3}\)
= lim \(\frac{\sqrt[3]{8+\frac{1}{n^3}}-1}{2-\frac{3}{n}}=\frac{2-1}{2}=\frac{1}{2}\)
Đề đúng là +2 trên tử phải nằm trong căn đầu tiên, nếu ko giới hạn sẽ là dương vô cùng
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x^2+x-2}{\sqrt{x^2+x+2}+2}+\frac{8-\left(7x+1\right)}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{x-1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{1}{4+2\sqrt[3]{7x+1}+\sqrt[3]{\left(7x+1\right)^2}}}{\sqrt{2}}\)
\(=\frac{\frac{3}{4}-\frac{1}{4+4+4}}{\sqrt{2}}=\frac{2}{3\sqrt{2}}=\frac{\sqrt{2}}{3}+0\)
\(\Rightarrow a+b+c=1+3+0=4\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{3x+1}-2}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3x+1-4}{\sqrt{3x+1}+2}\cdot\dfrac{1}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3x-3}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{3}{\left(x+1\right)\left(\sqrt{3x+1}+2\right)}=\dfrac{3}{\left(1+1\right)\left(\sqrt{3+1}+2\right)}\)
\(=\dfrac{3}{2\cdot4}=\dfrac{3}{8}\)
=>a=3;b=8
=>a2+b=9+8=17
\(\lim\dfrac{\sqrt{9n^2-n+1}}{4n-2}=\lim\dfrac{n\sqrt{9-\dfrac{1}{n}+\dfrac{1}{n^2}}}{n\left(4-\dfrac{2}{n}\right)}=\lim\dfrac{\sqrt{9-\dfrac{1}{n}+\dfrac{1}{n^2}}}{4-\dfrac{2}{n}}=\dfrac{3}{4}\)
\(\Rightarrow b-a=1\)