GP: gold point

SP: silver point

\(\omega=\omega_R=2\pi f_o=2\pi.2=4\pi\left(\dfrac{rad}{s}\right)\)

\(\overrightarrow{AB}.\overrightarrow{AC}=AB.ACcos\widehat{BAC}=a.a\sqrt{2}.cos45^0=a^2\)

\(\lim\limits_{x\rightarrow3^-}\sqrt{x^2+8x+3}=\sqrt{3^2+8.3+3}=6\)

\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)\Leftrightarrow\lim\limits_{x\rightarrow1^+}\dfrac{x+m}{x}=\lim\limits_{x\rightarrow1^-}\left(x^2-x+3\right)\\ \Leftrightarrow m+1=3\Leftrightarrow m=2\)

\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt[3]{x^3+x^2}-x\right)\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{x^2}{\left(\sqrt[3]{x^3+x^2}\right)^2+x\sqrt[3]{x^3+x^2}+x^2}\\ =\lim\limits_{x\rightarrow-\infty}\dfrac{1}{\left(\sqrt[3]{1+\dfrac{1}{x}}\right)^2+\sqrt[3]{1+\dfrac{1}{x}}+1}=\dfrac{1}{3}\)

@camcon: Ở bước cuối:

lim(tử) < 0; lim(mẫu) = 0; phải bổ sung thêm là (mẫu) < 0 nữa.

Lúc này kết quả cuối cùng sẽ là \(+\infty\), không phải \(-\infty\).

@camcon: \(\lim\limits_{x\rightarrow0^+}\dfrac{2x}{\sqrt{4x^2+x^3}}=1;\lim\limits_{x\rightarrow0^-}\dfrac{2x}{\sqrt{4x^2+x^3}}=-1\)

@camcon: Bước cuối không đúng.