Đáp án: $ L=-\dfrac52$
Giải thích các bước giải:
Ta có:
$L=\lim_{x\to1}\dfrac{\sqrt{4x-3}+\sqrt{2x-1}-3x+1}{(x-1)^2}$
$\to L=\lim_{x\to1}\dfrac{(\sqrt{4x-3}-(2x-1))+(\sqrt{2x-1}-x)}{(x-1)^2}$
$\to L=\lim_{x\to1}\dfrac{\dfrac{4x-3-(2x-1)^2}{\sqrt{4x-3}+(2x-1)}+\dfrac{2x-1-x^2}{\sqrt{2x-1}+x}}{(x-1)^2}$
$\to L=\lim_{x\to1}\dfrac{\dfrac{-4x^2+8x-4}{\sqrt{4x-3}+(2x-1)}+\dfrac{2x-1-x^2}{\sqrt{2x-1}+x}}{(x-1)^2}$
$\to L=\lim_{x\to1}\dfrac{-4}{\sqrt{4x-3}+(2x-1)}+\dfrac{-1}{\sqrt{2x-1}+x}$
$\to L=\dfrac{-4}{\sqrt{4\cdot 1-3}+(2\cdot 1-1)}+\dfrac{-1}{\sqrt{2\cdot 1-1}+1}$
$\to L=-\dfrac52$
