1a.
\(\lim\limits_{x\rightarrow-\infty}\left(x^3+2x^2-1\right)=\lim\limits_{x\rightarrow-\infty}x^3\left(1+\dfrac{2}{x}-\dfrac{1}{x^3}\right)=-\infty.1=-\infty\)
b. \(\lim\limits_{x\rightarrow-\infty}\left(4x^6+5x^2+1\right)=\lim\limits_{x\rightarrow-\infty}x^6\left(4+\dfrac{5}{x^4}+\dfrac{1}{x^6}\right)=+\infty.4=+\infty\)
c. \(\lim\limits_{x\rightarrow1}\dfrac{2x^2-5x+3}{x-1}=\lim\limits_{x\rightarrow1}\dfrac{\left(x-1\right)\left(2x-3\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(2x-3\right)=-1\)
d. \(\lim\limits_{x\rightarrow-5}\dfrac{3-\sqrt{4-x}}{x+5}=\lim\limits_{x\rightarrow-5}\dfrac{x+5}{\left(x+5\right)\left(3+\sqrt{4-x}\right)}=\lim\limits_{x\rightarrow-5}\dfrac{1}{3+\sqrt{4-x}}=\dfrac{1}{6}\)
2.
a. \(y'=\dfrac{\left(2x+2\right)\left(x+2\right)-\left(x^2+2x-3\right)}{\left(x+2\right)^2}=\dfrac{x^2+4x+7}{\left(x+2\right)^2}\)
b. \(y'=-4x\sqrt{1+2x^2}+\dfrac{2x\left(1-2x^2\right)}{\sqrt{1+2x^2}}=-\dfrac{12x^3+2x}{\sqrt{1+2x^2}}\)
c. \(y'=-14x^6+\dfrac{1}{2\sqrt{x}}\)
3.
Ta có: \(\left\{{}\begin{matrix}SB\perp\left(ABC\right)\Rightarrow SB\perp AC\\AC\perp AB\end{matrix}\right.\) \(\Rightarrow AC\perp\left(SAB\right)\)
Do \(\left\{{}\begin{matrix}BH\in\left(SAB\right)\\AC\perp\left(SAB\right)\end{matrix}\right.\) \(\Rightarrow BH\perp AC\)
Mà \(BH\perp SA\Rightarrow BH\perp\left(SAC\right)\)
