\(lim_{x->0}\frac{x.sin2x}{1-cos2x}\)
\(lim_{x->0}\frac{\sqrt{1-x}-1}{x}\)
\(lim_{x->0-}\frac{1}{x}\left(\frac{1}{x+1}-1\right)\)
\(lim_{x->0-}\frac{2x+\sqrt{-x}}{5x-\sqrt{-x}}\)
\(lim_{x->1}\frac{\sqrt[3]{6x-5}-\sqrt{4x-3}}{\left(x-1\right)^2}\)
l\(lim_{x->0}\left(1-x\right)tan\frac{\pi x}{2}\)
Câu dưới là 1 giới hạn hoàn toàn bình thường (không phải dạng vô định), bạn cứ thay số vào là được thôi
\(\lim\limits_{x\rightarrow0}\left(1-x\right)tan\frac{\pi x}{2}=\left(1-0\right).tan0=1\)
giai cau duoi thoi nha
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,\(Lim_{x->0}\frac{\ln\left(1+x\right)-xe^3}{x\tan\left(2x\right)}\)
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)\(lim_{x->2\frac{x-\sqrt{x+2}}{\sqrt{4x+1}-3}}\)
\(\lim_{x \to 0}\left ( \frac{e^x}{e^x-1}-\frac{1}{x} \right )\)
2 cách:
C1: Xài VCB tương đương khi x ->0
\(\lim\limits_{x\rightarrow0}\left(\dfrac{e^x}{e^x-1}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\left(\dfrac{e^x-1+1}{e^x-1}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\left(\dfrac{x+1}{x}-\dfrac{1}{x}\right)=\lim\limits_{x\rightarrow0}\dfrac{x}{x}=1\)
C2: Xài L'Hospital
\(=\lim\limits_{x\rightarrow0}\dfrac{e^x.x-e^x+1}{x.e^x-x}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x+e^x-e^x}{e^x.x+e^x-1}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x}{e^x.x+e^x-1}=\lim\limits_{x\rightarrow0}\dfrac{e^x.x+e^x}{e^x.x+2e^x-1}=1\)
\(lim_{x->1}\frac{\sqrt{6-2x}-\sqrt{x^2+3}}{\left(x-1\right)^2}\)
\(lim_{x->1^-}=\dfrac{2x+1}{x-1}\)
\(lim_{x->6}=\dfrac{\left(5x-4\right)\sqrt{2x-3}+x-84}{x-6}\)
Lời giải:
\(\lim\limits_{x\to 1-}\frac{2x+1}{x-1}=-\infty\) do với $x\to 1-$ thì $\lim(2x+1)=3>0$ và $\lim (x-1)=0$ và $x-1<0$
\(\lim\limits_{x\to 6}\frac{(5x-4)\sqrt{2x-3}+x-84}{x-6}=\lim\limits_{x\to 6}\frac{(5x-4)(\sqrt{2x-3}-3)+16(x-6)}{x-6}\)
\(=\lim\limits_{x\to 6}\frac{(5x-4).\frac{2(x-6)}{\sqrt{2x-3}+3}+16(x-6)}{x-6}=\lim\limits_{x\to 6}[\frac{2(5x-4)}{\sqrt{2x-3}+3}+16]=\frac{74}{3}\)
\(lim_{x\rightarrow1^-}\frac{\sqrt{x^2-x+3}}{2\left|x\right|-1}\)
\( \displaystyle\lim_{ x \rightarrow 0 } \left( \dfrac{ \sqrt[ 3 ]{ x+1 \phantom{\tiny{!}}} - \sqrt{ 1-x \phantom{\tiny{!}}} }{ x } \right) \)
\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{x+1}-1+1-\sqrt[]{1-x}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{x}{\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{x+1}+1}+\dfrac{x}{1+\sqrt[]{1-x}}}{x}\)
\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt[3]{\left(x+1\right)^3}+\sqrt[3]{x+1}+1}+\dfrac{1}{1+\sqrt[]{1-x}}\right)=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)