Tính $\lim_{x\rightarrow 0}\frac{1-\sqrt{2x^{2}+1}}{1-cosx}$
$\lim_{x\rightarrow 0}\frac{1-\sqrt{2x^{2}+1}}{1-cosx}$
#1
Đã gửi 26-03-2013 - 22:26
#2
Đã gửi 26-03-2013 - 23:01
$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt {2{x^2} + 1} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \sqrt {2{x^2} + 1} } \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)\left( {1 + \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^2}\left( {1 + \cos x} \right)}}{{\left( {1 - {{\cos }^2}x} \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^2}\left( {1 + \cos x} \right)}}{{{{\sin }^2}x\left( {1 + \sqrt {2{x^2} + 1} } \right)}}$
$\mathop {\lim }\limits_{x \to 0} \frac{{ - 2\left( {1 + \cos x} \right)}}{{\frac{{{{\sin }^2}x}}{{{x^2}}}\left( {1 + \sqrt {2{x^2} + 1} } \right)}} = \frac{{ - 4}}{2} = - 2$
Vì:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$
- VNSTaipro và phanquockhanh thích
#3
Đã gửi 28-03-2013 - 09:13
Tính $\lim_{x\rightarrow 0}\frac{1-\sqrt{2x^{2}+1}}{1-cosx}$
$\mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt {2{x^2} + 1} }}{{1 - \cos x}} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \sqrt {2{x^2} + 1} } \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)\left( {1 + \cos x} \right)}}{{\left( {1 - \cos x} \right)\left( {1 + \cos x} \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)}}$
$ = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^2}\left( {1 + \cos x} \right)}}{{\left( {1 - {{\cos }^2}x} \right)\left( {1 + \sqrt {2{x^2} + 1} } \right)}} = \mathop {\lim }\limits_{x \to 0} \frac{{ - 2{x^2}\left( {1 + \cos x} \right)}}{{{{\sin }^2}x\left( {1 + \sqrt {2{x^2} + 1} } \right)}}$
$\mathop {\lim }\limits_{x \to 0} \frac{{ - 2\left( {1 + \cos x} \right)}}{{\frac{{{{\sin }^2}x}}{{{x^2}}}\left( {1 + \sqrt {2{x^2} + 1} } \right)}} = \frac{{ - 4}}{2} = - 2$
Vì:
$\mathop {\lim }\limits_{x \to 0} \frac{{\sin x}}{x} = 1$
Ngắn hơn 1 chút: $\lim_{x\to 0}\frac{1-\sqrt{2x^{2}+1}}{1-cosx}=\lim_{x\to0}\frac{-2x^{2}}{2sin^{2}\frac{x}{2}(1+\sqrt{2x^{2}+1})}$
- phanquockhanh yêu thích
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