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\(\lim\limits_{x\rightarrow1}\frac{x^{2016}+x-2}{\sqrt{2018x+1}-\sqrt{x+2018}}=\lim\limits_{x\rightarrow1}\frac{2016x^{2015}+1}{\frac{1009}{\sqrt{2018x+1}}-\frac{1}{2\sqrt{x+2018}}}=\frac{2017}{\frac{1009}{\sqrt{2019}}-\frac{1}{2\sqrt{2019}}}=2\sqrt{2019}\)
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.
\(\lim\limits_{x\rightarrow1}\frac{\sqrt{x^2+x+2}-2+2-\sqrt[3]{7x+1}}{\sqrt{2}\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}-\frac{7\left(x-1\right)}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}}{\sqrt{2}\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{1}{\sqrt{2}}\left(\frac{x+2}{\sqrt{x^2+x+2}+2}-\frac{7}{\sqrt[3]{\left(7x+1\right)^2}+2\sqrt[3]{7x+1}+4}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\frac{3}{4}-\frac{7}{12}\right)=\frac{\sqrt{2}}{12}\)
\(\Rightarrow a+b+c=1+12+0=13\)
\(a=\lim\limits_{x\rightarrow a}\frac{\left(\sqrt{x}-\sqrt{a}\right)\left(x+\sqrt{ax}+a\right)}{\sqrt{x}-\sqrt{a}}=\lim\limits_{x\rightarrow a}\left(x+\sqrt{ax}+a\right)=3a\)
\(b=\lim\limits_{x\rightarrow1}\frac{x^{\frac{1}{n}}-1}{x^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow1}\frac{\frac{1}{n}x^{\frac{1-n}{n}}}{\frac{1}{m}x^{\frac{1-m}{m}}}=\frac{\frac{1}{n}}{\frac{1}{m}}=\frac{m}{n}\)
Ta có:
\(\lim\limits_{x\rightarrow1}\frac{1-\sqrt[n]{x}}{1-x}=\lim\limits_{x\rightarrow1}\frac{1-x^{\frac{1}{n}}}{1-x}=\lim\limits_{x\rightarrow1}\frac{-\frac{1}{n}x^{\frac{1-n}{n}}}{-1}=\frac{1}{n}\)
\(\Rightarrow c=\lim\limits_{x\rightarrow1}\frac{\left(1-\sqrt{x}\right)}{1-x}.\frac{\left(1-\sqrt[3]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[4]{x}\right)}{\left(1-x\right)}.\frac{\left(1-\sqrt[5]{x}\right)}{\left(1-x\right)}=\frac{1}{2}.\frac{1}{3}.\frac{1}{4}.\frac{1}{5}=\frac{1}{120}\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\frac{1}{x\sqrt{x}}}}+1}=\frac{1}{2}\)
\(e=\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{x}{\sqrt{1+x}+1}+\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\frac{1}{\sqrt{1+x}+1}+\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}\right)=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(f=\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{8x+11}-3+3-\sqrt{x+7}}{\left(x-1\right)\left(x-2\right)}=\lim\limits_{x\rightarrow2}\frac{\frac{8\left(x-2\right)}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{x-2}{3+\sqrt{x+7}}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow2}\frac{\frac{8}{\sqrt[3]{\left(8x+11\right)^2}+3\sqrt[3]{8x+11}+9}-\frac{1}{3+\sqrt{x+7}}}{x-1}=\frac{8}{27}-\frac{1}{6}=\frac{7}{54}\)
\(g=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{3x-2}-1+1-\sqrt{2x-1}}{\left(x-1\right)\left(x^2+x+1\right)}=\lim\limits_{x\rightarrow1}\frac{\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2\left(x-1\right)}{1+\sqrt{2x-1}}}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{\frac{3}{\sqrt[3]{\left(3x-2\right)^2}+\sqrt[3]{3x-2}+1}-\frac{2}{1+\sqrt{2x-1}}}{x^2+x+1}=0\)
\(h=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+9}+\sqrt[3]{2x-6}}{x^3+1}=\frac{\sqrt[3]{10}-\sqrt[3]{4}}{2}\)
Bài 1:
\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\frac{\sqrt{x+3}-2+2-\sqrt[3]{3x+5}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt{x+3}+2}-\frac{3\left(x-1\right)}{4+2\sqrt[3]{3x+5}+\sqrt[3]{\left(3x+5\right)^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt{x+3}+2}-\frac{3}{4+2\sqrt[3]{3x+5}+\sqrt[3]{\left(3x+5\right)^2}}\right)=0\)
\(f\left(1\right)=a+1\)
Để hàm số liên tục trên \([-3;+\infty)\Leftrightarrow\) hàm số liên tục tại \(x=1\)
\(\Leftrightarrow\lim\limits_{x\rightarrow1}f\left(x\right)=f\left(1\right)\Rightarrow a+1=0\Rightarrow a=-1\)
Bài 2:
Các hàm số đã cho đều liên tục trên R nên liên tục trên từng khoảng bất kì
a/ Xét \(f\left(x\right)=m\left(x-1\right)^3\left(x+2\right)+2x+3\)
\(f\left(-2\right)=-1\) ; \(f\left(1\right)=5\)
\(\Rightarrow f\left(-2\right).f\left(1\right)< 0;\forall m\)
\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-2;1\right)\) với mọi m
b/ \(m\left(sin^3x-cosx\right)=0\)
Nếu \(m=0\) pt có vô số nghiệm (thỏa mãn)
Nếu \(m\ne0\Leftrightarrow f\left(x\right)=sin^3x-cosx=0\)
\(f\left(0\right)=-1\) ; \(f\left(\frac{\pi}{2}\right)=1\)
\(\Rightarrow f\left(0\right).f\left(\frac{\pi}{2}\right)< 0\)
\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;\frac{\pi}{2}\right)\)
Phương trình luôn có nghiệm với mọi m
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)