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Question

Tìm giới hạn: $\mathop {\lim }\limits_{x \to 0}$ $\frac{\sqrt{5x^2+4}-2(x+1) }{\sin2x}$

Trần Nhật Tân · Answer 1 · 3/19/2013 10:29:24 PM

Xét hai trường hợp :
1.
$L=\mathop {\lim }\limits_{x \to 0^+}\frac{\sqrt{5x^2+4}-2(x+1) }{\sin2x}$
$L=\mathop {\lim }\limits_{x \to 0^+}\frac{2x }{\sin2x}.\mathop {\lim }\limits_{x \to 0^+}\frac{\sqrt{5x^2+4}-2(x+1) }{2x}$
$L=1.\mathop {\lim }\limits_{x \to 0^+}\left ( \sqrt{\dfrac{5}{4}+\dfrac{1}{x^2}}-2+\dfrac{1}{2x} \right )$
$L=+\infty.$
2.
$L=\mathop {\lim }\limits_{x \to 0^-}\frac{\sqrt{5x^2+4}-2(x+1) }{\sin2x}$
$L=\mathop {\lim }\limits_{x \to 0^-}\frac{2x }{\sin2x}.\mathop {\lim }\limits_{x \to 0^-}\frac{\sqrt{5x^2+4}-2(x+1) }{2x}$
$L=1.\mathop {\lim }\limits_{x \to 0^-}\left (- \sqrt{\dfrac{5}{4}+\dfrac{1}{x^2}}-2+\dfrac{1}{2x} \right )$
$L=-\infty.$

Trả lời 19-03-13 10:29 PM