lim\(\frac{\sqrt{1+\text{ax}}.\sqrt{1+bx}.\sqrt{1+cx}-1}{x}\)
x->0
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lim ( x ----> 0 ) \(\frac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{x}\)
= lim ( x----> 0 ) \(\frac{\sqrt[m]{1+ax}-1+1-\sqrt[n]{1+bx}}{x}\)
= lim ( x ---> 0 ) \(\frac{\sqrt[m]{1+ax}-1}{x}\)- lim ( x ---> 0 ) \(\frac{\sqrt[n]{1+bx}-1}{x}\)
= lim ( x ----> 0 ) \(\frac{ax}{x\left(\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1\right)}\)
- lim ( x ----> 0 ) \(\frac{bx}{x\left(\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1\right)}\)
= lim ( x -----> 0 ) \(\frac{a}{\sqrt[m]{\left(1+ax\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
- lim ( x ---> 0 ) \(\frac{b}{\sqrt[n]{\left(1+bx\right)^{n-1}}+\sqrt[n]{\left(1+bx\right)^{n-2}}+...+1}\)
= \(\frac{a}{m}-\frac{b}{n}\)
\(\frac{\sqrt{ax+1}\left(\sqrt[3]{bx+1}-1\right)+\sqrt{ax+1}-1}{x}=\frac{\frac{bx\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{ax}{\sqrt{ax+1}+1}}{x}=\frac{b\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{a}{\sqrt{ax+1}+1}\)
\(\Rightarrow\lim\limits_{x\rightarrow0}f\left(x\right)=a+b\Rightarrow a+b=1\)
\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
Mình sử dụng L'Hopital nhé, 2 loại căn thế này tìm liên hợp kép dài lắm :D
\(\lim\limits_{x\rightarrow1}\frac{\left(6x-5\right)^{\frac{1}{3}}-\left(4x-3\right)^{\frac{1}{2}}}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{2\left(6x-5\right)^{-\frac{2}{3}}-2\left(4x-3\right)^{-\frac{1}{2}}}{2\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-4\left(6x-5\right)^{-\frac{5}{3}}+2\left(4x-3\right)^{-\frac{3}{2}}}{1}=-2\)
Nếu ko bạn tách liên hợp như vầy:
\(\frac{\left(\sqrt[3]{6x-5}-2x+1\right)+\left(2x-1-\sqrt{4x-3}\right)}{\left(x-1\right)^2}\)
Sẽ khử được \(\left(x-1\right)^2\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{1}=\frac{a}{n}\)
\(B=\lim\limits_{x\rightarrow0}\frac{\left(1+ax\right)^{\frac{1}{n}}-1}{\left(1+bx\right)^{\frac{1}{m}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{a}{n}\left(1+ax\right)^{\frac{1-n}{n}}}{\frac{b}{m}\left(1+bx\right)^{\frac{1-m}{m}}}=\frac{am}{bn}\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+bx}\sqrt[4]{1+cx}\left(\sqrt{1+ax}-1\right)+\sqrt[4]{1+cx}\left(\sqrt[3]{1+bx}-1\right)+\left(\sqrt[4]{1+cx}-1\right)}{x}\)
\(C=\lim\limits_{x\rightarrow0}\sqrt[3]{1+bx}\sqrt[4]{1+cx}.\frac{\sqrt{1+ax}-1}{x}+\lim\limits_{x\rightarrow0}\sqrt[4]{1+cx}.\frac{\sqrt[3]{1+bx}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{1+cx}-1}{x}\)
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:
\(\lim\limits_{x\rightarrow a}\frac{f\left(x\right)}{g\left(x\right)}=\lim\limits_{x\rightarrow a}\frac{f'\left(x\right)}{g'\left(x\right)}\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(x+1\right)^{\frac{1}{3}}-1}{\left(2x+1\right)^{\frac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\frac{1}{2}\left(2x+1\right)^{-\frac{3}{4}}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}\)
\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}=\frac{3\sqrt{5}}{0}=+\infty\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(3x+1\right)\left(4x+1\right)}\left(\sqrt{2x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}\left(\sqrt{3x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}-1}{x}\)
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 \(\frac{0}{0}\)
Tui nghĩ cái này L'Hospital chứ giải thông thường là ko ổn :)
\(M=\lim\limits_{x\rightarrow0}\dfrac{\left(1+4x\right)^{\dfrac{1}{2}}-\left(1+6x\right)^{\dfrac{1}{3}}}{1-\cos3x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{2}\left(1+4x\right)^{-\dfrac{1}{2}}.4-\dfrac{1}{3}\left(1+6x\right)^{-\dfrac{2}{3}}.6}{3.\sin3x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{-\dfrac{1}{4}.4\left(1+4x\right)^{-\dfrac{3}{2}}.4+\dfrac{2}{9}.6.6\left(1+6x\right)^{-\dfrac{5}{3}}}{3.3.\cos3x}\)
Giờ thay x vô là được
\(N=\lim\limits_{x\rightarrow0}\dfrac{\left(1+ax\right)^{\dfrac{1}{m}}-\left(1+bx\right)^{\dfrac{1}{n}}}{\left(1+x\right)^{\dfrac{1}{2}}-1}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{m}.\left(1+ax\right)^{\dfrac{1}{m}-1}.a-\dfrac{1}{n}\left(1+bx\right)^{\dfrac{1}{n}-1}.b}{\dfrac{1}{2}\left(1+x\right)^{-\dfrac{1}{2}}}=\dfrac{\dfrac{a}{m}-\dfrac{b}{n}}{\dfrac{1}{2}}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\left(1+2x\right)^{\dfrac{1}{2}}-\left(1+3x\right)^{\dfrac{1}{3}}}=\lim\limits_{x\rightarrow0}\dfrac{n\left(1+mx\right)^{n-1}.m-m\left(1+nx\right)^{m-1}.n}{\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{1}{2}}.2-\dfrac{1}{3}\left(1+3x\right)^{-\dfrac{2}{3}}.3}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{n\left(n-1\right)\left(1+mx\right)^{n-2}.m-m\left(m-1\right)\left(1+nx\right)^{m-2}.n}{-\dfrac{1}{2}\left(1+2x\right)^{-\dfrac{3}{2}}.2+\dfrac{2}{9}.3.3\left(1+3x\right)^{-\dfrac{5}{3}}}=....\left(thay-x-vo-la-duoc\right)\)
1/ \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-2+2-\sqrt[3]{8-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2x}{\sqrt{1+x}+1}+\dfrac{x}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{2}{\sqrt{1+x}+1}+\dfrac{1}{4+2\sqrt[3]{8-x}+\sqrt[3]{\left(8-x\right)^2}}\right)=\dfrac{13}{12}\)
2/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-2-\left(\sqrt{x+3}-2\right)}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x-1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{x-1}{\sqrt{x+3}+2}}{\left(x-1\right)\left(x-2\right)}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{1}{\sqrt[3]{\left(x+7\right)^2}+2\sqrt[3]{x+7}+4}-\dfrac{1}{\sqrt{x+3}+2}}{x-2}=\dfrac{1}{6}\)
3/ \(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}=\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-2+2-\sqrt{5-x^2}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{\left(x^2-1\right)}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{x^2-1}{2+\sqrt{5-x^2}}}{x^2-1}\)
\(=\lim\limits_{x\rightarrow1}\left(\dfrac{1}{\sqrt[3]{\left(x^2+7\right)^2}+2\sqrt[3]{x^2+7}+4}+\dfrac{1}{2+\sqrt{5-x^2}}\right)=\dfrac{1}{3}\)
4/ \(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}=\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-3-\left(\sqrt[3]{8x+43}-3\right)}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{x+2}{\sqrt{x+11}+3}-\dfrac{8\left(x+2\right)}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{\left(2x-1\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{\dfrac{1}{\sqrt{x+11}+3}-\dfrac{8}{\sqrt[3]{\left(8x+43\right)^2}+3\sqrt[3]{8x+43}+9}}{2x-1}=\dfrac{7}{270}\)
5/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1-\left(\sqrt[m]{1+bx}-1\right)}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{bx}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{a}{\sqrt[n]{\left(1+ax\right)^{n-1}}+\sqrt[n]{\left(1+ax\right)^{n-2}}+...+1}-\dfrac{b}{\sqrt[m]{\left(1+bx\right)^{m-1}}+\sqrt[m]{\left(1+ax\right)^{m-2}}+...+1}\)
\(=\dfrac{a}{n}-\dfrac{b}{m}\)
6/ \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-\sqrt{1+4x}+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\left(\sqrt[3]{1+6x}-1\right)+\sqrt{1+4x}-1}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\dfrac{6x}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4x}{\sqrt{1+4x}+1}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{6\sqrt{1+4x}}{\sqrt[3]{\left(1+6x\right)^2}+\sqrt[3]{1+6x}+1}+\dfrac{4}{\sqrt{1+4x}+1}\right)=4\)