Edited by vantho302, 01-11-2012 - 09:17.
\[\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {a^{\sin x}}}}{{{x^3}}}\]
Started By xuanphu2602, 01-11-2012 - 07:08
#1
Posted 01-11-2012 - 07:08
xuanphu2602
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Lính mới
\[\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {a^{\sin x}}}}{{{x^3}}}\]
#2
Posted 04-11-2012 - 22:39
cuong148
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Hạ sĩ
Áp dụng liên tiếp L'Hospital 3 lần.(Có vẻ trâu bò
)
${{\lim }_{x\to 0}}\ln a.\frac{{{a}^{x}}-\cos x.{{a}^{\sin x}}}{3.{{x}^{2}}}={{\lim }_{x\to 0}}\frac{\ln a}{3.2}.\frac{\ln a.{{a}^{x}}+\sin x.{{a}^{\sin x}}-{{(\cos x)}^{2}}.\ln a.{{a}^{\sin x}}}{x}={{\lim }_{x\to 0}}\frac{\ln a}{6}.\frac{\ln a.\ln a.{{a}^{x}}+\sin x.\cos x.\ln a.{{a}^{\sin x}}+\ln a.\sin 2x.{{a}^{\sin x}}-\sin x.{{\cos }^{2}}x.\ln a.{{a}^{\sin x}}}{1}=\frac{{{\ln }^{3}}a}{6}$
)${{\lim }_{x\to 0}}\ln a.\frac{{{a}^{x}}-\cos x.{{a}^{\sin x}}}{3.{{x}^{2}}}={{\lim }_{x\to 0}}\frac{\ln a}{3.2}.\frac{\ln a.{{a}^{x}}+\sin x.{{a}^{\sin x}}-{{(\cos x)}^{2}}.\ln a.{{a}^{\sin x}}}{x}={{\lim }_{x\to 0}}\frac{\ln a}{6}.\frac{\ln a.\ln a.{{a}^{x}}+\sin x.\cos x.\ln a.{{a}^{\sin x}}+\ln a.\sin 2x.{{a}^{\sin x}}-\sin x.{{\cos }^{2}}x.\ln a.{{a}^{\sin x}}}{1}=\frac{{{\ln }^{3}}a}{6}$
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