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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 ) \(\frac{b}{\sqrt[n]{\left(1+bx\right)^{n-1}}+\sqrt[n]{\left(1+bx\right)^{n-2}}+...+1}\)
= \(\frac{a}{m}-\frac{b}{n}\)
\(A=\lim\limits_{x\rightarrow0}\frac{\left(x+1\right)^{\frac{1}{3}}-1}{\left(2x+1\right)^{\frac{1}{4}}-1}=\lim\limits_{x\rightarrow0}\frac{\frac{1}{3}\left(x+1\right)^{-\frac{2}{3}}}{\frac{1}{2}\left(2x+1\right)^{-\frac{3}{4}}}=\frac{\frac{1}{3}}{\frac{1}{2}}=\frac{2}{3}\)
\(B=\lim\limits_{x\rightarrow7}\frac{\sqrt[3]{4x-1}\sqrt{x-2}}{\sqrt[4]{2x+2}-2}=\frac{3\sqrt{5}}{0}=+\infty\)
\(C=\lim\limits_{x\rightarrow0}\frac{\sqrt{\left(3x+1\right)\left(4x+1\right)}\left(\sqrt{2x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}\left(\sqrt{3x+1}-1\right)}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{4x+1}-1}{x}\)
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 \(\frac{0}{0}\)
\(\lim\limits_{x\rightarrow0}\frac{\left(2x+1\right)^{\frac{1}{2}}-\left(3x+1\right)^{\frac{1}{3}}}{x^2}=\lim\limits_{x\rightarrow0}\frac{\left(2x+1\right)^{-\frac{1}{2}}-\left(3x+1\right)^{-\frac{2}{3}}}{2x}\)
\(=\lim\limits_{x\rightarrow0}\frac{-\left(2x+1\right)^{-\frac{3}{2}}+2\left(3x+1\right)^{-\frac{5}{3}}}{2}=\frac{-1+2}{2}=\frac{1}{2}\)
Vậy nó ko phải dạng vô định, cứ thay số trực tiếp
\(=\frac{2}{0}=+\infty\)
Nếu là mũ 3 thì nó là dạng 0/0 rút gọn được. Nên chắc là đề ghi nhầm đấy
\(\dfrac{\sqrt{1+2x}\sqrt[3]{1+3x}\sqrt[4]{1+4x}-1}{x}\)
\(=\dfrac{\sqrt[3]{1+3x}\sqrt[4]{1+4x}\left(\sqrt{1+2x}-1\right)}{x}+\dfrac{\sqrt[4]{1+4x}\left(\sqrt[3]{1+3x}-1\right)}{x}+\dfrac{\sqrt[4]{1+4x}+1}{x}\)
Dùng L'Hopital dễ dàng chứng minh với mọi n nguyên dương ta có:
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+nx}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(1+nx\right)^{\dfrac{1}{n}}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{1}{n}n\left(1+nx\right)^{\dfrac{1-n}{n}}}{1}=\dfrac{n}{n}=1\)
\(\Rightarrow\) giới hạn đã cho bằng \(\sqrt[3]{1+3.0}\sqrt[4]{1+4.0}.1+\sqrt[4]{1+4.0}.1+1=1+1+1=3\)
\(a=\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{3x+1}-\sqrt{x+3}\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\lim\limits_{x\rightarrow1}\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{2}{\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\frac{2}{2.4}=\frac{1}{4}\)
\(b=\frac{3}{0}=+\infty\)
\(c=\frac{-13}{0}=-\infty\)
Bạn tự hiểu là giới hạn tiến đến đâu nhé, làm biếng gõ đủ công thức
a. \(\frac{\sqrt{1+x}-1+1-\sqrt[3]{1+x}}{x}=\frac{\frac{x}{\sqrt{1+x}+1}-\frac{x}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}}{x}=\frac{1}{\sqrt{1+x}+1}-\frac{1}{1+\sqrt[3]{1+x}+\sqrt[3]{\left(1+x\right)^2}}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}\)
b.
\(\frac{1-x^3-1+x}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1-x\right)\left(1+x\right)}{\left(1-x\right)^2\left(1+x+x^2\right)}=\frac{x\left(1+x\right)}{\left(1-x\right)\left(1+x+x^2\right)}=\frac{2}{0}=\infty\)
c.
\(=\frac{-2}{\sqrt[3]{\left(2x-1\right)^2}+\sqrt[3]{\left(2x+1\right)^2}+\sqrt[3]{\left(2x-1\right)\left(2x+1\right)}}=\frac{-2}{\infty}=0\)
d.
\(=x\sqrt[3]{3-\frac{1}{x^3}}-x\sqrt{1+\frac{2}{x^2}}=x\left(\sqrt[3]{3-\frac{1}{x^3}}-\sqrt{1+\frac{2}{x^2}}\right)=-\infty\)
e.
\(=\frac{2x^2-8x+8}{\left(x-1\right)\left(x-2\right)\left(x-2\right)\left(x-3\right)}=\frac{2\left(x-2\right)^2}{\left(x-1\right)\left(x-3\right)\left(x-2\right)^2}=\frac{2}{\left(x-1\right)\left(x-3\right)}=\frac{2}{-1}=-2\)
f.
\(=\frac{2x}{x\sqrt{4+x}}=\frac{2}{\sqrt{4+x}}=1\)