Giải thích các bước giải:
1.Ta có:
$\lim_{x\to1}\dfrac{\sqrt{5x-4}-\sqrt{2x-1}}{x-1}$
$=\lim_{x\to1}\dfrac{(\sqrt{5x-4}-1)+(1-\sqrt{2x-1})}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{5x-4-1}{\sqrt{5x-4}+1}+\dfrac{1-(2x-1)}{1+\sqrt{2x-1}}}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{5(x-1)}{\sqrt{5x-4}+1}+\dfrac{2(1-x)}{1+\sqrt{2x-1}}}{x-1}$
$=\lim_{x\to1}\dfrac{5}{\sqrt{5x-4}+1}-\dfrac{2}{1+\sqrt{2x-1}}$
$=\dfrac{5}{\sqrt{5\cdot 1-4}+1}-\dfrac{2}{1+\sqrt{2\cdot 1-1}}$
$=\dfrac32$
2.Ta có:
$\lim_{x\to 1}\dfrac{\sqrt{5x-4}-1}{x-1}$
$=\lim_{x\to 1}\dfrac{\dfrac{5x-4-1}{\sqrt{5x-4}+1}}{x-1}$
$=\lim_{x\to 1}\dfrac{\dfrac{5(x-1)}{\sqrt{5x-4}+1}}{x-1}$
$=\lim_{x\to 1}\dfrac{5}{\sqrt{5x-4}+1}$
$=\dfrac{5}{\sqrt{5\cdot 1-4}+1}$
$=\dfrac52$
3.Ta có:
$\lim_{x\to1}\dfrac{\sqrt{2x-1}-1}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{2x-1-1}{\sqrt{2x-1}+1}}{x-1}$
$=\lim_{x\to1}\dfrac{\dfrac{2(x-1)}{\sqrt{2x-1}+1}}{x-1}$
$=\lim_{x\to1}\dfrac{2}{\sqrt{2x-1}+1}$
$=\dfrac{2}{\sqrt{2\cdot 1-1}+1}$
$=1$
