Question

\(lim_{x->1^-}=\dfrac{2x+1}{x-1}\)

\(lim_{x->6}=\dfrac{\left(5x-4\right)\sqrt{2x-3}+x-84}{x-6}\)

 

Answer 1

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 $x-1<0$

\(\lim\limits_{x\to 6}\frac{(5x-4)\sqrt{2x-3}+x-84}{x-6}=\lim\limits_{x\to 6}\frac{(5x-4)(\sqrt{2x-3}-3)+16(x-6)}{x-6}\)

\(=\lim\limits_{x\to 6}\frac{(5x-4).\frac{2(x-6)}{\sqrt{2x-3}+3}+16(x-6)}{x-6}=\lim\limits_{x\to 6}[\frac{2(5x-4)}{\sqrt{2x-3}+3}+16]=\frac{74}{3}\)