[Toán 11] Tìm giới hạn

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niemkieuloveahbu

[TEX]lim_{x->0} \frac{(\sqrt{1+tanx} - \sqrt{1+sinx})}{{x^3}}[/TEX]
[TEX]=lim_{x \to 0} \frac{tanx-sinx}{cosx.x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}[/TEX]
[TEX]=lim_{x \to 0} \frac{2sinxsin^2\frac{x}{2}}{cosx.x^3(\sqrt{1+tanx}+\sqrt{1+sinx})}[/TEX]
[TEX]=lim_{x \to 0} \frac{2sinxsin^2\frac{x}{2}}{4x.(\frac{x}{2})^2.cosx.(\sqrt{1+tanx}+\sqrt{1+sinx})}[/TEX]
[TEX]= lim_{x \to 0}\frac{1}{2cosx(\sqrt{1+tanx}+ \sqrt{1+sinx})}=\frac{1}{4}[/TEX]
 
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