giới hạn

Question

tìm giới hạn A=

$\mathop {\lim }\limits_{x \to 1} \frac{\sqrt{2x-1}+ \sqrt{5-4x}-2x^{2}}{x^{2}-1}$

Dép Lê Con Nhà Quê · Answer 1 · 9/22/2013 9:17:36 PM

√+5−4x−−−−−√−2x2x2−1

tìm giới hạn A=

limx→12x−1−−−−−√+5−4x−−−−−√−2x2x2−1

Ta có

$\dfrac{(\sqrt{2x-1}-1) +(\sqrt{5-4x}-1) -2(x^2-1)}{x^2-1}= \dfrac{(\sqrt{2x-1}-1)}{x^2-1} + \dfrac{(\sqrt{5-4x}-1)}{x^2-1} -2$

$=\dfrac{2(x-1)}{(x-1)(x+1)(\sqrt{2x-1}+1)} -\dfrac{4(x-1)}{(x-1)(x+1)(\sqrt{5-4x}+1)}-2$

$=\dfrac{2}{(x+1)(\sqrt{2x-1}+1)} -\dfrac{4}{(x+1)(\sqrt{5-4x}+1)}-2 =A$

Vậy

$\lim \limits_{x\to 1} A = \dfrac{1}{2}-1-2=-\dfrac{5}{2}$

√+5−4x−−−−−√−2x2x2−1

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