Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$ 
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*Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$   

*If $0<\theta < \frac{\pi}{2} $ and $\sin 2\theta=\cos 3\theta~~$ then find the value of $\sin\theta$


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*$\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}$
$=\displaystyle\lim_{x\rightarrow 0}\frac{5^x-1-(4^x-1)}{x}$
$=\displaystyle\lim_{x\rightarrow 0}\frac{5^x-1}{x}$ -$\displaystyle\lim_{x\rightarrow 0}\frac{4^x-1}{x}$
$=\log_e5-\log_e4  ~~~~~~$  $[\because\displaystyle\lim_{x\rightarrow 0}\frac{a^x-1}{x}=\log_a (a>0)]$
$=\log_e(\frac{5}{4})$  

