Giải giúp mình với ạ 1 6: Cho $\lim_{x\rightarrow2}\frac{2x^2-5x+2}{x^3-8}=a,~\lim_{x\rightarrow3}\frac{\sqrt{2x+3}-3}{x^2-4x+3}=b.$,Các mệnh đề sau đúng hay sai?,$a)~a=\frac14.$ $\lim_{x\rightarrow2}\frac{2x^2-5x+2}{x^3-2}=\lim\frac{(x-2)(2x^2-x)}{(x-2)(x^2+2x+4)}$,$b)~b=6.$,$c)~4a+6b=2.$ $=\lim\frac{2x^2-x}{x^2+2x+4}=$,d) l,a,b là dãy số tăng.

Question

Accepted Answer

$\displaystyle \begin{array}{{>{\displaystyle}l}}
\lim _{x ightarrow 2}\frac{2x^{2} -5x+2}{x^{3} -8} =\lim _{x ightarrow 2}\frac{( 2x-1)( x-2)}{( x-2)\left( x^{2} +2x+4 ight)} =\lim _{x ightarrow 2}\frac{2x-1}{x^{2} +2x+4} =\frac{1}{4}\
\lim _{x ightarrow 3}\frac{\sqrt{2x+3} -3}{x^{2} -4x+3} =\frac{2x-6}{\left(\sqrt{2x+3} +3 ight)( x-3)( x-1)} =\frac{2}{\left(\sqrt{2x+3} +3 ight)( x-1)} \ =\ \frac{1}{6}
\end{array}$
⟹ a) đúng
b) sai
c) đúng
d) sai