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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.
\(y'=4\left(x-5\right)^3\)
\(\lim\limits_{x\rightarrow-\infty}\frac{17}{x^2+4}=\frac{17}{+\infty}=0\)
\(\lim\limits_{x\rightarrow-\infty}\frac{-2x^2+x-1}{x^2+3}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}-\frac{1}{x^2}}{1+\frac{3}{x^2}}=\frac{-2+0+0}{1+0}=-2\)
\(\lim\limits_{x\rightarrow-\infty}\frac{x+\sqrt{4x^2-1}}{-3x+2}=\lim\limits_{x\rightarrow-\infty}\frac{x-x\sqrt{4-\frac{1}{x^2}}}{x\left(-3+\frac{2}{x}\right)}=\lim\limits_{x\rightarrow-\infty}\frac{1-\sqrt{4-\frac{1}{x^2}}}{-3+\frac{3}{x}}=\frac{1-2}{-3}=\frac{1}{3}\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}{\left(x-1\right)\left(x^2+x-1\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x+1\right)\left(x^2+1\right)}{x^2+x-1}=\frac{4}{1}=4\)
\(b=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(c=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)^2}{\left(x^2+1\right)\left(x^2-9\right)}=\lim\limits_{x\rightarrow3}\frac{\left(x+1\right)\left(x-3\right)}{\left(x^2+1\right)\left(x+3\right)}=\frac{0}{60}=0\)
\(d=\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{x^2-2x+1}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2x-2}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=10\)
\(e=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(f=\lim\limits_{x\rightarrow-2}\frac{\left(x+2\right)\left(x-2\right)\left(x^2+4\right)}{\left(x+2\right)x^2}=\lim\limits_{x\rightarrow-2}\frac{\left(x-2\right)\left(x^2+4\right)}{x^2}=-8\)
Hai câu d, e khai triển thì dài quá nên làm biếng sử dụng L'Hopital
\(A=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{1}=\frac{a}{n}\)
\(B=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{\left(1+bx\right)^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{\frac{b}{m}\left(1+bx\right)^{\frac{1-m}{m}}}=\frac{am}{bn}\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+bx}\sqrt[4]{1+cx}\left(\sqrt{1+ax}-1\right)+\sqrt[4]{1+cx}\left(\sqrt[3]{1+bx}-1\right)+\left(\sqrt[4]{1+cx}-1\right)}{x}\)
\(C=\lim\limits_{x\rightarrow0}\sqrt[3]{1+bx}\sqrt[4]{1+cx}.\frac{\sqrt{1+ax}-1}{x}+\lim\limits_{x\rightarrow0}\sqrt[4]{1+cx}.\frac{\sqrt[3]{1+bx}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{1+cx}-1}{x}\)
Từ câu A ta có: \(\lim\limits_{x\rightarrow0}\frac{\sqrt[n]{1+ax}-1}{x}=\frac{a}{n}\)
\(\Rightarrow C=\frac{a}{2}+\frac{b}{3}+\frac{c}{4}\)
Bạn sử dụng định lý L'Hopital cho giới hạn vô định:
\(\lim\limits_{x\rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\lim\limits_{x\rightarrow a}\frac{f'\left(x\right)}{g'\left(x\right)}\)
Bạn tự hiểu là giới hạn khi x tiến tới 1 nhé
a/\(=lim\frac{\left(x-1\right)\left(x^{2015}+x^{2014}+...+x+1\right)}{\left(x-1\right)\left(x^{2014}+x^{2013}+...+x+1\right)}=lim\frac{x^{2015}+x^{2014}+...+x+1}{x^{2014}+x^{2013}+...+x+1}=\frac{2016}{2015}\)
b/ \(=lim\frac{\left(x-1\right)\left(x^{m-1}+x^{m-2}+...+x+1\right)}{\left(x-1\right)\left(x^{n-1}+x^{n-2}+...+x+1\right)}=lim\frac{x^{m-1}+...+1}{x^{n-1}+...+1}=\frac{m}{n}\)
Hoặc nếu bạn được sử dụng L'Hopital thì cứ việc đạo hàm tử-mẫu, lẹ hơn các trên nhiều
\(a=\lim\limits_{x\rightarrow0}\frac{x^2}{x\left(\sqrt{1+x^2}+1\right)}=\lim\limits_{x\rightarrow0}\frac{x}{\sqrt{1+x^2}+1}=\frac{0}{2}=0\)
\(b=\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-2+2-\sqrt{5-x^2}}{x-1}=\lim\limits_{x\rightarrow1}\frac{\frac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{\left(x-1\right)\left(x+1\right)}{2+\sqrt{5-x^2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\frac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}+\frac{x+1}{2+\sqrt{5-x^2}}\right)=\frac{1}{12}+\frac{1}{2}=\frac{7}{12}\)
\(c=\lim\limits_{x\rightarrow0}\frac{2x}{x\left(\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}\right)}=\lim\limits_{x\rightarrow0}\frac{2}{\sqrt[3]{\left(1+x\right)^2}+\sqrt[3]{\left(1+x\right)\left(1-x\right)}+\sqrt[3]{\left(1-x\right)^2}}=\frac{2}{3}\)
\(d=\frac{\sqrt[3]{6}}{0}=+\infty\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(x+1\right)^{\frac{1}{3}}-1}{\left(2x+1\right)^{\frac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\frac{1}{2}\left(2x+1\right)^{-\frac{3}{4}}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}\)
\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}=\frac{3\sqrt{5}}{0}=+\infty\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(3x+1\right)\left(4x+1\right)}\left(\sqrt{2x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}\left(\sqrt{3x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}-1}{x}\)
Xét \(\lim\limits_{x\rightarrow0}\frac{\sqrt{ax+1}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\left(ax+1\right)^{\frac{1}{2}}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{2}\left(ax+1\right)^{-\frac{1}{2}}}{1}=\frac{a}{2}\)
\(\Rightarrow C=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}=\frac{9}{2}\)
\(D=\lim\limits_{x\rightarrow0}\frac{\left(1+4x\right)^{\frac{1}{2}}-\left(1+6x\right)^{\frac{1}{3}}}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\left(1+4x\right)^{-\frac{1}{2}}-2\left(1+6x\right)^{-\frac{2}{3}}}{2x}\)
\(D=\lim\limits_{x\rightarrow0}\frac{-2\left(1+4x\right)^{-\frac{3}{2}}+4\left(1+6x\right)^{-\frac{5}{3}}}{1}=-2+4=2\)
\(E=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-\left(1+bx\right)^{\frac{1}{n}}}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}-\frac{b}{n}\left(1+bx\right)^{\frac{1-n}{n}}}{1}=\frac{a-b}{n}\)
Vì câu đó ko phải dạng vô định, nó là 1 giới hạn bình thường.
Mình đoán bạn ghi nhầm đề, đề bài là \(\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}-\sqrt{x+2}}{\sqrt[4]{2x+2}-2}\) thì hợp lý hơn, đây là 1 giới hạn vô định \(\frac{0}{0}\)
