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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) \(t = {e^x} - 1 \Leftrightarrow {e^x} = t + 1 \Leftrightarrow x = \ln \left( {t + 1} \right)\)
\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - 1}}{x} = \mathop {\lim }\limits_{t \to 0} \frac{t}{{\ln \left( {t + 1} \right)}} = 1\)
Đổi biến \(\cos x=y^{20}\). Khi \(x\rightarrow0\) thì \(y\rightarrow0\). Ta có :
\(L=\lim\limits_{y\rightarrow0}\frac{y^5-y^4}{1-y^{40}}=-\lim\limits_{y\rightarrow0}\frac{y^4\left(y-1\right)}{y^{40}-1}\)
\(=-\lim\limits_{y\rightarrow0}\frac{y-1}{\left(y-1\right)\left(y^{39}+y^{38}+.....+y+1\right)}=-\frac{1}{40}\)
Lời giải:
1.
\(\lim\limits_{x\to -1}\frac{x^{2019}+1}{x^2+x}=\lim\limits_{x\to -1}\frac{(x+1)(x^{2018}-x^{2017}+x^{2016}-....-x+1)}{x(x+1)}=\lim\limits_{x\to -1}\frac{x^{2018}-x^{2017}+x^{2016}-....-x+1}{x}\)
\(=\frac{(-1)^{2018}-(-1)^{2017}+(-1)^{2016}+....-(-1)+1}{-1}\)
\(=\frac{\underbrace{1+1+....+1+1}_{2019}}{-1}=\frac{2019}{-1}=-2019\)
2.
\(\lim\limits_{x\to 1}\frac{(x-1)+(x^2-1)+(x^3-1)+....+(x^n-1)}{x-1}\\ =\lim\limits_{x\to 1}\frac{(x-1)+(x-1)(x+1)+(x-1)(x^2+x+1)+....+(x-1)(x^{n-1}+x^{n-2}+...+x+1)}{x-1}\)
$\lim\limits_{x\to 1}[1+(x+1)+(x^2+x+1)+....+(x^{n-1}+x^{n-2}+...+x+1)]$
$=1+2+3+....+n=n(n+1):2$
\(\)
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 : \(\lim\limits_{x\rightarrow0}\frac{1-\cos ax}{x^2}=\lim\limits_{x\rightarrow0}\frac{2\sin^2\frac{ax}{2}}{x^2}=\lim\limits_{x\rightarrow0}\left[\left(\frac{\sin\frac{ax}{2}.\sin\frac{ax}{2}}{\frac{ax}{2}.\frac{ax}{2}}\right).\frac{a^2}{2}\right]\)
\(=\frac{a^2}{2}\left(\lim\limits_{y\rightarrow0}\frac{\sin y}{y}\right)^2=\frac{a^2}{2}\)
1: \(\lim\limits_{x\rightarrow4}\dfrac{1-x}{\left(x-4\right)^2}=-\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow4}1-x=1-4=-3< 0\\\lim\limits_{x\rightarrow4}\left(x-4\right)^2=\left(4-4\right)^2=0\end{matrix}\right.\)
2: \(\lim\limits_{x\rightarrow3^+}\dfrac{2x-1}{x-3}=+\infty\)
vì \(\left\{{}\begin{matrix}\lim\limits_{x\rightarrow3^+}2x-1=2\cdot3-1=5>0\\\lim\limits_{x\rightarrow3^+}x-3=3-3>0\end{matrix}\right.\) và x-3>0
3: \(\lim\limits_{x\rightarrow2^+}\dfrac{-2x+1}{x+2}\)
\(=\dfrac{-2\cdot2+1}{2+2}=\dfrac{-3}{4}\)
4: \(\lim\limits_{x\rightarrow1^-}\dfrac{3x-1}{x+1}=\dfrac{3\cdot1-1}{1+1}=\dfrac{2}{2}=1\)
Lại xài L'Hopital:
\(=\lim\limits_{x\rightarrow1}\dfrac{1+2x+3x^2+...+nx^{n-1}}{1+2x+3x^2+...+mx^{m-1}}=\dfrac{1+2+...+n}{1+2+...+m}=\dfrac{n\left(n+1\right)}{m\left(m+1\right)}\)
Để giới hạn đã cho hữu hạn
\(\Rightarrow\sqrt{x^2+mx-m-3}-x=0\) có nghiệm \(x=4\)
\(\Rightarrow\sqrt{16+4m-m-3}-4=0\)
\(\Rightarrow\sqrt{3m+13}=4\Rightarrow m=1\)
Khi đó:
\(\lim\limits_{x\rightarrow4}\dfrac{\sqrt{x^2+x-4}-x}{x^2-5x+4}=\lim\limits_{x\rightarrow4}\dfrac{x-4}{\left(x-1\right)\left(x-4\right)\left(\sqrt{x^2+x-4}+x\right)}\)
\(=\lim\limits_{x\rightarrow4}\dfrac{1}{\left(x-1\right)\left(\sqrt{x^2+x-4}+x\right)}=\dfrac{1}{3\left(\sqrt{4^2+4-4}+4\right)}=\dfrac{1}{24}\)
a) \(lim\dfrac{-2n+1}{n}=lim\dfrac{\dfrac{-2n}{n}+\dfrac{1}{n}}{\dfrac{n}{n}}=lim\dfrac{-2+\dfrac{1}{n}}{1}=\dfrac{lim\left(-2\right)+\dfrac{lim1}{n}}{lim1}=\dfrac{-2+0}{1}=-\dfrac{2}{1}=-2\)
b) \(\lim\limits_{x\rightarrow1}\dfrac{3-\sqrt{x+8}}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{9-\left(x+8\right)}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{x-1}{\left(x-1\right)\left(3+\sqrt{x+8}\right)}=\lim\limits_{x\rightarrow1}\dfrac{1}{3+\sqrt{x+8}}=\dfrac{1}{3+\sqrt{1+8}}=\dfrac{1}{3+3}=\dfrac{1}{9}\)
