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\(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x\)
Ta có : \(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x}{1+x}\right)^x=\lim\limits_{x\rightarrow+\infty}\left(1-\frac{1}{1+x}\right)^x\)
Đặt \(-\frac{1}{1+x}=\frac{1}{t}\Rightarrow\begin{cases}x=-\left(1+t\right)\\x\rightarrow+\infty;t\rightarrow-\infty\end{cases}\)
\(\Rightarrow L=\lim\limits_{t\rightarrow-\infty}\left(1+\frac{1}{t}\right)^{-\left(1+t\right)}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)^{1+t}}=\lim\limits_{t\rightarrow-\infty}\frac{1}{\left(1+\frac{1}{t}\right)\left(1+\frac{1}{t}\right)^t}=\frac{1}{1.e}=\frac{1}{e}\)
\(L=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{2x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+x^3\right)}{x^3.\frac{2}{x^2}}=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+x^3\right)}{x^3}.\frac{x^3}{2}\right]=1.0=0\)
\(L=\lim\limits_{x\rightarrow0}\frac{\ln x-1}{\tan x}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{\frac{\sin x}{\cos x}}=\lim\limits_{x\rightarrow0}\frac{\ln\left(1+2x\right)}{2x.\frac{\sin x}{x}.\frac{1}{2\cos x}}\)
\(=\lim\limits_{x\rightarrow0}\left[\frac{\ln\left(1+2x\right)}{2x}.\frac{1}{\frac{\sin x}{x}}.2\cos x\right]=1.\frac{1}{1}.2.1=2\)
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
Đặt \(t=x-e\Rightarrow\begin{cases}x=t+e\\x\rightarrow e;t\rightarrow0\end{cases}\)
\(\Rightarrow L=\lim\limits_{t\rightarrow0}\frac{\ln\left(t+e\right)-\ln e}{t}=\lim\limits_{t\rightarrow0}\frac{\ln\left(\frac{t+e}{e}\right)}{t}=\lim\limits_{t\rightarrow0}\left[\frac{\ln\left(1+\frac{t}{e}\right)}{\frac{t}{e}}\right]=\frac{1}{e}\)
Quan sát đồ thị ta thấy x → -∞ thì f(x) → 0; khi x → 3- thì f(x) → -∞;
khi x → -3+ thì f(x) x → +∞.
b) f(x) =
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f(x) =
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= -∞ vì
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> 0 và
= -∞.
f(x) =
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.
= +∞
vì
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> 0 và
= +∞.
Bài 2:
\(\lim\limits_{x\to 2}\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}=\lim\limits_{x\to 2}\frac{x^2-x-2}{(x+\sqrt{x+2}).\frac{4x+1-9}{\sqrt{4x+1}+3}}=\lim\limits_{x\to 2}\frac{(x-2)(x+1)(\sqrt{4x+1}+3)}{(x+\sqrt{x+2}).4(x-2)}=\lim\limits_{x\to 2}\frac{(x+1)(\sqrt{4x+1}+3)}{4(x+\sqrt{x+2})}=\frac{9}{8}\)
Bài 3:
\(\lim\limits_{x\to 0-}\frac{1-\sqrt[3]{x-1}}{x}=-\infty \)
\(\lim\limits_{x\to 0+}\frac{1-\sqrt[3]{x-1}}{x}=+\infty \)
Bài 4:
\(\lim\limits_{x\to -\infty}\frac{x^2-5x+1}{x^2-2}=\lim\limits_{x\to -\infty}\frac{1-\frac{5}{x}+\frac{1}{x^2}}{1-\frac{2}{x^2}}=1\)
Bài 5:
\(\lim\limits_{x\to +\infty}\frac{2x^2-4}{x^3+3x^2-9}=\lim\limits_{x\to +\infty}\frac{\frac{2}{x}-\frac{4}{x^3}}{1+\frac{3}{x}-\frac{9}{x^3}}=0\)
Bài 6:
\(\lim\limits_{x\to 2- }\frac{2x-1}{x-2}=\lim\limits_{x\to 2-}\frac{2(x-2)+3}{x-2}=\lim\limits_{x\to 2-}\left(2+\frac{3}{x-2}\right)=-\infty \)
Bài 7:
\(\lim\limits _{x\to 3+ }\frac{8+x-x^2}{x-3}=\lim\limits _{x\to 3+}\frac{1}{x-3}.\lim\limits _{x\to 3+}(8+x-x^2)=2(+\infty)=+\infty \)
Bài 8:
\(\lim\limits _{x\to -\infty}(8+4x-x^3)=\lim\limits _{x\to -\infty}(-x^3)=+\infty \)
Bài 9:
\(\lim\limits _{x\to -1}\frac{\sqrt[3]{x}+1}{\sqrt{x^2+3}-2}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{x^2+3-4}=\lim\limits _{x\to -1}\frac{x+1}{\sqrt[3]{x^2}-\sqrt[3]{x}+1}.\frac{\sqrt{x^2+3}+2}{(x-1)(x+1)}\)
\(\lim\limits _{x\to -1}\frac{\sqrt{x^2+3}+2}{(\sqrt[3]{x^2}-\sqrt[3]{x}+1)(x-1)}=\frac{-2}{3}\)
\(L=\lim\limits_{x\rightarrow+\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}=\lim\limits_{x\rightarrow+\infty}\left(1+\frac{3}{x-2}\right)^{2x-1}\)
Đặt \(\begin{cases}\frac{3}{x-2}=\frac{1}{t}\Rightarrow x=3t+2\\x\rightarrow+\infty;t\rightarrow+\infty\end{cases}\)
\(L=\lim\limits_{x\rightarrow+\infty}\left(1+\frac{1}{t}\right)^{6t+3}=\lim\limits_{x\rightarrow+\infty}\left\{\left[\left(1+\frac{1}{t}\right)^t\right]^6.\left(1+\frac{1}{t}\right)^3\right\}=e^6.1^3=e^6\)