dễ thấy hàm số có dạng 0/0 áp dùng l'hospital
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-1}{x}=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt[n]{1+ax}-1\right)'}{\left(x\right)'}\\ =\lim\limits_{x\rightarrow0}\dfrac{a}{n\sqrt[n]{\left(1+ax\right)^{n-1}}}=\dfrac{a}{n}\)
Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)
1) \(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
2)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x+7}-\sqrt{x+3}}{x^2-3x+2}\)
3)\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[3]{x^2+7}-\sqrt{5-x^2}}{x^2-1}\)
4)\(\lim\limits_{x\rightarrow-2}\dfrac{\sqrt{x+11}-\sqrt[3]{8x+43}}{2x^2+3x-2}\)
5) \(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[n]{1+ax}-\sqrt[m]{1+bx}}{x}\)
6)\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}.\sqrt[3]{1+6x}-1}{x}\)
Tìm các giới hạn sau :
A=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[n]{1+ax}-1}{x}\left(n\in N^{\cdot},a\ne0\right)\)
B=\(\lim\limits_{x\rightarrow0}\frac{\sqrt[n]{1+ax}-1}{\sqrt[m]{1+bx}-1}\) với\(ab\ne0\)
C=\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+\alpha x}\sqrt[3]{1+\beta x}\sqrt[4]{1+\gamma x}-1}{x}\) với\(\alpha\beta\gamma\ne0\)
Tính các giới hạn sau:\(I_1=\lim\limits_{x\rightarrow1}\dfrac{\left(1-\sqrt{x}\right)\left(1-\sqrt[3]{x}\right)....\left(1-\sqrt[n]{x}\right)}{\left(1-x\right)^{n-1}}\)
\(I_2=\lim\limits_{x\rightarrow0}\dfrac{\left(\sqrt{1+x^2}+x\right)^n-\left(\sqrt{1+x^2}-x\right)^n}{x}\)
Cho biết \(\lim\limits_{x\rightarrow0}\dfrac{sinax}{ax}=1\left(a\ne0\right)\). Tìm \(\lim\limits_{x\rightarrow0}\dfrac{1-cos2017x}{x^2}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{2x+3}-x}{x^2-4x+3}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1}{\sqrt[4]{2x+1}-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{x^2}\)
tính \(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{4x^2+1}-\sqrt{x+5}}{2x-7}\)
help pls
\(\lim\limits_{x\rightarrow1^+}\dfrac{x^2-x+1}{x^2-1}\)
\(\lim\limits_{x\rightarrow0}\dfrac{2\sqrt{1+x}-\sqrt[3]{8-x}}{x}\)
\(\lim\limits_{x\rightarrow3}\dfrac{\sqrt{x+6}-3}{\sqrt{2x-2}-2}\)
\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{4x+1}-\sqrt[3]{2x+1}}{x}\)
\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt{4x+5}-3}{\sqrt[3]{5x+3}-2}\)
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt[4]{2x+3}+\sqrt[3]{2+3x}}{\sqrt{x+2}-1}\)
