a) Sử dụng phép đổi biến \(t = \frac{1}{x},\) tìm giới hạn \(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}}.\)
b) Với \(y = {\left( {1 + x} \right)^{\frac{1}{x}}},\) tính ln y và tìm giới hạn của \(\mathop {\lim }\limits_{x \to 0} \ln y.\)
c) Đặt \(t = {e^x} - 1.\) Tính x theo t và tìm giới hạn \(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x}.\)
a) Ta có \(t = \frac{1}{x},\) nên khi x tiến đến 0 thì t tiến đến dương vô cùng do đó
\(\mathop {\lim }\limits_{x \to 0} {\left( {1 + x} \right)^{\frac{1}{x}}} = \mathop {\lim }\limits_{t \to + \infty } {\left( {1 + \frac{1}{t}} \right)^t} = e\)
b) \(\ln y = \ln {\left( {1 + x} \right)^{\frac{1}{x}}} = \frac{1}{x}\ln \left( {1 + x} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \ln y = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + x} \right)}}{x} = 1\)
c) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)
Tìm giới hạn : \(L=\lim\limits_{x\rightarrow0}\frac{\sqrt[4]{\cos x}-\sqrt[5]{\cos x}}{\sin^2x}\)
Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :
\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)
\(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)
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. \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}\)
1. Ta có : \(\lim\limits_{x\rightarrow0}\frac{\tan ax}{\tan bx}=\lim\limits_{x\rightarrow0}\left(\frac{\sin ax}{\sin bx}.\frac{\cos ax}{\cos bx}\right)=\lim\limits_{x\rightarrow0}\frac{\sin ax}{\sin bx}=\lim\limits_{x\rightarrow0}\left(\frac{\frac{\sin ax}{ax}}{\frac{\sin bx}{bx}}.\frac{ax}{bx}\right)=\frac{a}{b}\frac{\lim\limits_{x\rightarrow0}\frac{\sin ax}{ax}}{\lim\limits_{x\rightarrow0}\frac{\sin bx}{bx}}=\frac{a}{b}\frac{\lim\limits_{y\rightarrow0}\frac{\sin y}{y}}{\lim\limits_{z\rightarrow0}\frac{\sin z}{z}}=\frac{a}{b}\)
2. Ta có : \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\sin^2\frac{ax}{2}}{x^2}=\lim\limits_{x\rightarrow0}\left[\left(\frac{\sin\frac{ax}{2}.\sin\frac{ax}{2}}{\frac{ax}{2}.\frac{ax}{2}}\right).\frac{a^2}{2}\right]\)
\(=\frac{a^2}{2}\left(\lim\limits_{y\rightarrow0}\frac{\sin y}{y}\right)^2=\frac{a^2}{2}\)
Tìm giới hạn: \(A=\lim\limits_{x\rightarrow\dfrac{1}{2}^-}\dfrac{4x+1}{1-2x}\)
\(=\lim\limits_{x\rightarrow\dfrac{1}{2}^-}\dfrac{4.\left(\dfrac{1}{2}\right)+1}{0}=+\infty\)
tính giới hạn lim(x→0)\(\dfrac{ }{\dfrac{2\sqrt{2x+1}-\sqrt[3]{x^2+x+8}}{x}}\)
=\(\dfrac{a}{b}\)
tính a-2b=?
\(=\lim\limits_{x\rightarrow0}\dfrac{2\left(\sqrt[]{2x+1}-1\right)+2-\sqrt[3]{x^2+x+8}}{x}\)
\(=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{2.2x}{\sqrt[]{2x+1}+1}-\dfrac{x\left(x+1\right)}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{4}{\sqrt[]{2x+1}+1}-\dfrac{x+1}{\sqrt[3]{\left(x^2+x+8\right)^2}+2\sqrt[3]{x^2+x+8}+4}\right)\)
\(=\dfrac{23}{12}\)
\(lim\frac{Sin2x}{1+cos^3x}\)khi x->\(\pi\)tính giới hạn trên !!!
giới hạn \(\lim\limits_{x\rightarrow3}\dfrac{x+1-\sqrt{5x+1}}{x-\sqrt{4x-3}}=\dfrac{a}{b}\). tìm a,b biết a/b tối giản
\(\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+2x+1-5x-1\right)\left(x+\sqrt{4x-3}\right)}{\left(x^2-4x+3\right)\left(x+1+\sqrt{5x+1}\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x-3\right)\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x-3\right)\left(x+1+\sqrt{5x+1}\right)}\)
\(=\lim\limits_{x\rightarrow3}\dfrac{x\left(x+\sqrt{4x-3}\right)}{\left(x-1\right)\left(x+1+\sqrt{5x+1}\right)}=\dfrac{9}{8}\)
Tìm giới hạn A = lim x → 1 x 3 - 3 x 2 + 2 x 2 - 4 x + 3 .
A. + ∞
B. - ∞
C. 3 2
D. 1
Tìm giới hạn :
\(\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan^3x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}\)
\(L=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan^3x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}=\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x\left(\tan^2x-3\right)}{\cos\left(x+\frac{\pi}{6}\right)}\)
\(=\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\left(\tan x-\sqrt{3}\right)\left(\tan x+\sqrt{3}\right)}{\sin\left(\frac{\pi}{3}-x\right)}=\sqrt{3}.2\sqrt{3}\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\tan x-\sqrt{3}}{\sin\left(\frac{\pi}{3}-x\right)}\)
\(=6\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{\sin\left(\frac{\pi}{3}-x\right)}{\cos x.\cos\frac{\pi}{3}\sin\left(\frac{\pi}{3}-x\right)}=-12\lim\limits_{x\rightarrow\frac{\pi}{3}}\frac{1}{\cos x}=-24\)
Tìm các giới hạn sau :
a, lim\(\dfrac{2x^2+x-6}{x^3+8}\) khi x→-2
b, lim\(\dfrac{x^4-x^2-72}{x^2-2x-3}\) khi x→3
c, lim\(\dfrac{x^5+1}{x^3+1}\) khi x→-1
d, lim \(\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)\) khi x→1
a) \(\lim\limits_{x\rightarrow-2}\dfrac{2x^2+x-6}{x^3+8}=\lim\limits_{x\rightarrow-2}\dfrac{\left(2x-3\right)\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\\ =\lim\limits_{x\rightarrow-2}\dfrac{2x-3}{x^2-2x+4}=-\dfrac{7}{12}\).
b) \(\lim\limits_{x\rightarrow3}\dfrac{x^4-x^2-72}{x^2-2x-3}=\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\\ =\lim\limits_{x\rightarrow3}\dfrac{\left(x^2+8\right)\left(x+3\right)}{x+1}=\dfrac{51}{2}\).
c) \(\lim\limits_{x\rightarrow-1}\dfrac{x^5+1}{x^3+1}=\lim\limits_{x\rightarrow-1}\dfrac{\left(x+1\right)\left(x^4-x^3+x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\lim\limits_{x\rightarrow-1}\dfrac{x^4-x^3+x^2-x+1}{x^2-x+1}=\dfrac{5}{3}\).
d) \(\lim\limits_{x\rightarrow1}\left(\dfrac{2}{x^2-1}-\dfrac{1}{x-1}\right)=\lim\limits_{x\rightarrow1}\left(\dfrac{2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\right)\\ =\lim\limits_{x\rightarrow1}\dfrac{1-x}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1}\dfrac{-1}{x+1}=-\dfrac{1}{2}\).
