\(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+ax}-1}{x}=\lim\limits_{x\rightarrow0}\frac{\left(\sqrt{1+ax}-1\right)\left(\sqrt{1+ax}+1\right)}{x\left(\sqrt{1+ax}+1\right)}\)
\(=\lim\limits_{x\rightarrow0}\frac{ax}{x\left(\sqrt{1+ax}+1\right)}=\lim\limits_{x\rightarrow0}\frac{a}{\sqrt{1+ax}+1}=\frac{a}{2}\)
\(\Rightarrow\frac{a}{2}=3\Rightarrow A=6\)
Cho biết \(\lim\limits_{x\rightarrow0}\dfrac{sinax}{ax}=1\left(a\ne0\right)\). Tìm \(\lim\limits_{x\rightarrow0}\dfrac{1-cos2017x}{x^2}\)
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\(\alpha\beta\gamma\ne0\)
Bài 1
a. \(\lim\limits_{x\rightarrow+\infty}\frac{1+2\sqrt{x}-x}{x+3}\) b. \(\lim\limits_{x\rightarrow+\infty}\frac{x^3+3x-1}{x^2\sqrt{x}+x}\) c. \(\lim\limits_{x\rightarrow-\infty}\frac{x+2\sqrt{1-x}}{1-x}\)
Bài 2: Tính các giới hạn sau biết \(\lim\limits_{x\rightarrow0}\frac{\sin x}{x}=1\)
a. \(\lim\limits_{x\rightarrow0}\frac{1-\cos x}{1-\cos3x}\) b. \(\lim\limits_{x\rightarrow0}\frac{\cot x-\sin x}{x^3}\) c. \(\lim\limits_{x\rightarrow\infty}\frac{x.\sin x}{2x^2}\)
a. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}\) f. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7-3}}{2-\sqrt{x+3}}\)
b. \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}\) g. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}\)
c. \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}\) h. \(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}\)
d. \(\lim\limits_{x\rightarrow1}\frac{3x-2\sqrt{4x^2-x-2}}{x^2-3x+2}\) k. \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)
e. \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}\)
Tính các giới hạn sau:\(M=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{1+4x}-\sqrt[3]{1+6x}}{1-cos3x}\)
\(N=\lim\limits_{X\rightarrow0}\dfrac{\sqrt[m]{1+ax}-\sqrt[n]{1+bx}}{\sqrt{1+x}-1}\)
\(V=\lim\limits_{x\rightarrow0}\dfrac{\left(1+mx\right)^n-\left(1+nx\right)^m}{\sqrt{1+2x}-\sqrt[3]{1+3x}}\)
tìm các giới hạn sau:
a, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+x^2}-1}{x}\)
b,\(\lim\limits_{x\rightarrow1}\frac{\sqrt[3]{x+7}-\sqrt{5-x^2}}{x-1}\)
c, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+x}-\sqrt[3]{1-x}}{x}\)
d, \(\lim\limits_{x\rightarrow2}\frac{\sqrt[3]{4x-2}}{x-2}\)
tìm các giới hạn sau:
a, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\)
b, \(\lim\limits_{x\rightarrow0}\frac{\sqrt[3]{1+x^2}-1}{x^2}\)
c, \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+2}+\sqrt{x+7}-5}{x-2}\)
tìm các giới hạn sau:
a, \(\lim\limits_{x\rightarrow-3}\frac{x+\sqrt{3-2x}}{x^2+3x}\)
b, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+9}+\sqrt{x+16}-7}{x}\)
c, \(\lim\limits_{x\rightarrow\frac{1}{2}}\frac{8x^2-1}{6x^2-5x+1}\)
d, \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{4-\sqrt{x^2+16}}\)
Cho a, b là hai số cho trước với \(b\ne0\), tìm các giới hạn sau :
1. \(\lim\limits_{x\rightarrow0}\frac{\tan ax}{\tan bx}\)
2. \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}\)
