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\(\lim\limits_{x\rightarrow0}\dfrac{\left(x^2+2014\right)\left(1-2014x\right)^{\dfrac{1}{2014}}-2014}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{2x\left(1-2014x\right)^{\dfrac{1}{2014}}+\left(x^2+2014\right).\dfrac{1}{2014}.\left(-2014\right).\left(1-2014x\right)^{-\dfrac{2013}{2014}}}{1}\)
\(=-2014\)
Bạn hướng dẫn mik chỗ dấu bằng thứ nhất được không ạ
Bài 1:
\(a=\lim\limits_{x\rightarrow-1}\frac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\lim\limits_{x\rightarrow-1}\frac{x^4-x^3+x^2-x+1}{x^2-x+1}=\frac{5}{3}\)
\(b=\frac{1-5+1}{0}=\frac{-3}{0}=-\infty\)
\(c=\lim\limits_{x\rightarrow1}\frac{x\left(1+2x\right)\left(1+3x\right)+2x\left(1+3x\right)+3x}{x}=\lim\limits_{x\rightarrow1}\left[\left(1+2x\right)\left(1+3x\right)+2\left(1+3x\right)+3\right]=1+2+3=6\)
\(d=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-1}{5x^4+2x}=\frac{4}{0}=+\infty\)
Bài 2:
\(a=\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(b=\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
\(c=\lim\limits_{x\rightarrow0}\frac{x+x^2+...+x^n-n}{x-1}=\frac{-n}{-1}=n\)
\(\left(1+x\right)\left(1+2x\right)...\left(1+nx\right)=x\left(1+2x\right)...\left(1+nx\right)+\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+\left(1+3x\right)...\left(1+nx\right)\)
\(=...\)
\(=x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx+1\)
\(\Rightarrow\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)...\left(1+nx\right)-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\frac{x\left(1+2x\right)...\left(1+nx\right)+2x\left(1+3x\right)...\left(1+nx\right)+...+nx}{x}\)
\(=\lim\limits_{x\rightarrow0}\left[\left(1+2x\right)...\left(1+nx\right)+2\left(1+3x\right)...\left(1+nx\right)+...+n\right]\)
\(=1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)
1.
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{x^2-3x-4}}{1-x^2}=\lim\limits_{x\to (-1)-}\frac{\sqrt{(x+1)(x-4)}}{(1-x)(1+x)}\)
\(=\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{(x-1)\sqrt{-(x+1)}}=-\infty\) do:
\(\lim\limits_{x\to (-1)-}\frac{\sqrt{4-x}}{x-1}=\frac{-\sqrt{5}}{2}<0\) và \(\lim\limits_{x\to (-1)-}\frac{1}{\sqrt{-(x+1)}}=+\infty\)
2.
\(\lim\limits_{x\to 2+}\left(\frac{1}{x-2}-\frac{x+1}{\sqrt{x+2}-2}\right)=\lim\limits_{x\to 2+}\frac{1-(x+1)(\sqrt{x+2}+2)}{x-2}=-\infty\) do:
\(\lim\limits_{x\to 2+}\frac{1}{x-2}=+\infty\) và \(\lim\limits_{x\to 2+}[1-(x+1)(\sqrt{x+2}+2)]=-11<0\)
\(a=\lim\limits_{x\rightarrow0}\frac{3x\left(1+x\right)\left(1+2x\right)}{x}+\lim\limits_{x\rightarrow0}\frac{2x\left(1+x\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)-1}{x}\)
\(=\lim\limits_{x\rightarrow0}3\left(1+x\right)\left(1+2x\right)+\lim\limits_{x\rightarrow0}2\left(1+x\right)+1=3+2+1=6\)
\(b=\lim\limits_{x\rightarrow0}\frac{\left(x^5+5x^4+10x^3+10x^2+5x+1\right)-\left(1+5x\right)}{x^5+x^2}\)
\(=\lim\limits_{x\rightarrow0}\frac{x^2\left(x^3+5x^2+10\right)}{x^2\left(x^3+1\right)}=\lim\limits_{x\rightarrow0}\frac{x^3+5x^2+10}{x^3+1}=10\)
tại sao a = \(\lim\limits_{x\rightarrow0}\frac{3x\left(1+x\right)\left(1+2x\right)}{x}\)
Bài 1:
a. \(\lim\limits_{x\rightarrow-1}\frac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\frac{5x^4}{3x^2}=\frac{5}{3}\)
b. \(\lim\limits_{x\rightarrow1}\frac{4x^6-5x^5+x}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{24x^5-25x^4+1}{2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\frac{120x^4-100x^3}{2}=\frac{120-100}{2}=10\)
c. \(\lim\limits_{x\rightarrow0}\frac{\left(1+2x\right)\left(1+3x\right)x}{x}+\lim\limits_{x\rightarrow0}\frac{\left(1+3x\right)2x}{x}+\lim\limits_{x\rightarrow0}\frac{3x+1-1}{x}=1+2+3=6\)
d. \(\lim\limits_{x\rightarrow0}\frac{\left(1+x\right)^5-\left(1+5x\right)}{x^5+x^2}=\lim\limits_{x\rightarrow0}\frac{5\left(1+x\right)^4-5}{5x^4+2x}\)
\(=\lim\limits_{x\rightarrow0}\frac{20\left(1+x\right)^3}{20x^3+2}=\frac{20}{2}=10\)
Bài 2:
\(\lim\limits_{x\rightarrow1}\frac{x^m-1}{x^n-1}=\lim\limits_{x\rightarrow1}\frac{mx^{m-1}}{nx^{n-1}}=\frac{m}{n}\)
\(\lim\limits_{x\rightarrow a}\frac{x-a}{x^n-a^n}=\lim\limits_{x\rightarrow a}\frac{1}{nx^{n-1}}=\frac{1}{n.a^{n-1}}\)
\(\lim\limits_{x\rightarrow1}\frac{x-x^2}{\left(2x-1\right)\left(x^5-3\right)}=\frac{1-1^2}{\left(2-1\right)\left(1-3\right)}=\frac{0}{-2}=0\)
\(\lim\limits_{x\rightarrow0}x\left(1-\frac{1}{x}\right)=\lim\limits_{x\rightarrow0}\left(x-1\right)=0-1=-1\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(x+1\right)^{\frac{1}{3}}-1}{\left(2x+1\right)^{\frac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\frac{1}{2}\left(2x+1\right)^{-\frac{3}{4}}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}\)
\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}=\frac{3\sqrt{5}}{0}=+\infty\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(3x+1\right)\left(4x+1\right)}\left(\sqrt{2x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}\left(\sqrt{3x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}-1}{x}\)
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 \(\frac{0}{0}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\left(1+2013x\right)^{2014}-\left(1-2014x\right)^{2013}}{x^2}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{2013.2014\left(1+2013x\right)^{2013}+2013.2014\left(1-2014x\right)^{2012}}{2x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{2013^3.2014\left(1+2013x\right)^{2012}-2012.2013.2014^2\left(1-2014x\right)^{2011}}{2}\)
\(=\dfrac{2013^3.2014-2012.2013.2014^2}{2}=...\)
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