Giới hạn nào sau đây tồn tại:
A, \(lim_{x\rightarrow+\infty}sin2x\) B, \(lim_{x\rightarrow+\infty}cos3x\) C, \(lim_{x\rightarrow0}sin\frac{1}{2x}\) D, \(lim_{x\rightarrow1}sin\frac{1}{2x}\)
Tìm các giới hạn sau:
a) \(lim_{x\rightarrow0}\dfrac{tan3x}{sin5x}\)
b) \(lim_{x\rightarrow0}\dfrac{cos2x-1}{sin^23x}\)
c) \(lim_{x\rightarrow1}\dfrac{x^2-4x+3}{sin\left(x-1\right)}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
Tính giới hạn
a, \(Lim_{n->+\infty}\frac{1+sin\left(n\right)+2^{n+2}}{2-2n+2^n}\)
b,\(Lim_{x->0}\frac{e^x-1-xcos\left(x\right)}{x\left(e^{2x}-1\right)}\)
c,\(Lim_{n->+\infty}\sqrt[2n]{8^n+9^n}\)
d,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
Sử dụng định nghĩa tính giới hạn $\lim\limits_{x\rightarrow +\infty} \dfrac2{3x+1}$.
tìm các giới hạn sau
a)\(lim_{x->-\infty}\left(3x+\sqrt{1-2x+9x^2}\right)\)
b)\(lim_{x->+\infty\left(x-\sqrt{1+x+x^2}\right)}\)
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. \(\lim\limits_{x\rightarrow\infty}\frac{x.\sin x}{2x^2}\)
Bài 1:
\(a=\lim\limits_{x\rightarrow+\infty}\frac{\frac{1}{x}+\frac{2}{\sqrt{x}}-1}{1+\frac{3}{x}}=-1\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{1+\frac{3}{x^2}-\frac{1}{x^3}}{\frac{1}{\sqrt{x}}+\frac{1}{x^2}}=\frac{1}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow-\infty}\frac{1-2\sqrt{\frac{1}{x^2}-\frac{1}{x}}}{\frac{1}{x}-1}=\frac{1}{-1}=-1\)
Bài 2:
\(a=\lim\limits_{x\rightarrow0}\frac{1-cosx}{1-cos3x}=\lim\limits_{x\rightarrow0}\frac{sinx}{3sin3x}=\lim\limits_{x\rightarrow0}\frac{\frac{sinx}{x}}{9.\frac{sin3x}{3x}}=\frac{1}{9}\)
\(b=\lim\limits_{x\rightarrow0}\frac{cotx-sinx}{x^3}=\frac{\infty}{0}=+\infty\)
\(c=\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}\)
Mà \(\left|sinx\right|\le1\Rightarrow\left|\frac{sinx}{2x}\right|\le\frac{1}{\left|2x\right|}\)
Mà \(\lim\limits_{x\rightarrow\infty}\frac{1}{2\left|x\right|}=0\Rightarrow\lim\limits_{x\rightarrow\infty}\frac{sinx}{2x}=0\)
\(\lim_{n\rightarrow\infty}\frac{1-2n+3n^3}{n^3+n}\) Tính giới hạn
\(=\lim_{n\rightarrow\infty}\frac{\frac{1}{n^3}-\frac{2n}{n^3}+\frac{3n^3}{n^3}}{\frac{n^3}{n^3}+\frac{n}{n^3}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{n^3}-\frac{2}{n^2}+3}{1+\frac{1}{n^2}}=\frac{0-0+3}{1+0}=3\)
Tìm giới hạn hàm số
a) \(\text{ }lim_{x->3\frac{\sqrt{2x^2-2x-3}-\sqrt{x^2+2x-6}}{x^2-4x+3}}\)
b)\(lim_{x->1\frac{x^3-x^2+2x-2}{x-1}}\)
c)\(lim_{x->1\frac{x^3-x^2+2x-2}{\sqrt{x}-1}}\)
d)\(lim_{x->2\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}}\)
\(lim_{x\rightarrow+\infty}\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{x+\sqrt{x+\sqrt{x}}-x}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)
\(=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\sqrt{\frac{1}{x}}}}{\sqrt{1+\sqrt{\frac{1}{x}+\sqrt{\frac{1}{x^3}}}}+1}=\frac{1}{1+1}=\frac{1}{2}\)