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#2
04 µØÅÒ¤Á 2013, 21:32
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Indeterminate form of type $1^\infty $. Transfrom using $$\lim_{x \to \ 0}\left(\,\frac{1}{cos(4x)}\right)^{\frac{3}{tan^2(5x)} }=exp\left(\lim_{x \to \ 0}\frac{3ln\left(\,\frac{1}{cos(4x)}\right) }{tan^2(5x)} \right)$$
Indeterminate form of type $0/0$. Applying L'Hospital's rule $$ \lim_{x \to \ 0}\frac{ln\left(\frac{1}{cos(4x)} \right) }{tan^2(5x)}$$ will get $$=exp\left(\lim_{x \to \ 0} \frac{6tan(4x)cos^2(5x)}{5tan(5x)} \right)$$ Indeterminate form of type $0/0$. Applying L'Hospital's rule again $$=exp\left(\lim_{x \to \ 0} \frac{24sec^2(4x)cos^2(5x)-60cos(5x)sin(5x)tan(4x)}{25sec^2(5x)} \right)=exp\left(\frac{24}{25} \right) $$ 
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