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

Question

1. Show that $\displaystyle\lim_{x\rightarrow 0}\frac{5^x-4^x}{x}=\log_e\left({\frac{5}{4}}\right)$
2. If $0<\theta < \frac{\pi}{2} $ and $\sin 2\theta=\cos 3\theta~~$ then find the value of $\sin\theta$

Answer 1

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

Answer 2

$$\sin2\theta=\cos3\theta$$

$$\implies 2\sin\theta\cos\theta=4\cos^3\theta-3\cos\theta--->(1)$$

As $0<\theta<\frac\pi2,\cos\theta\ne0\implies 2\sin\theta=4\cos^2\theta-3$ $=4(1-\sin^2\theta)-3=1-4\sin^2\theta$

$$\text{So, }4\sin^2\theta+2\sin\theta-1=0--->(2)$$

$$\implies \sin\theta=\frac{-2\pm\sqrt{4-4(4)(-1)}}{2\cdot4}=\frac{-1\pm\sqrt5}4$$

As $0<\theta<\frac\pi2,\sin\theta>0 \implies \sin\theta=\frac{-1+\sqrt5}4 $

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Determination of $\theta$:

$$\cos3\theta=\sin2\theta=\cos\left(\frac\pi2-2\theta\right)$$

$$\implies 3\theta=2n\pi\pm\left(\frac\pi2-2\theta\right)\text{ where } n \text{ is any integer}$$

Taking the '+' sign, $\theta=\frac{(4n+1)\pi}{10}$

$2\pi$ will divide $\frac{(4n_1+1)\pi}{10}-\frac{(4n_2+1)\pi}{10}=\frac{2(n_1-n_2)\pi}5$ if $5\mid (n_1-n_2)$

So, $5$ the in-congruent values of $n$ will give us $5$ the in-congruent values of $\theta$

Those values of $\theta$ are $\frac\pi{10},\frac{5\pi}{10}=\frac\pi2,\frac{9\pi}{10}=\pi-\frac{\pi}{10},\frac{13\pi}{10}=\pi+\frac{3\pi}{10},\frac{17\pi}{10}=2\pi-\frac{3\pi}{10}$

Taking the '-' sign, $\theta=2n\pi-\frac\pi2$ which clearly gives us exactly one in-congruent value of $\theta$, namely $-\frac\pi2$

Clearly, for $\cos\theta=0$ in $(1), \theta=\pm\frac\pi2$

Now, $\sin\frac{9\pi}{10}=\sin\left(\pi-\frac{\pi}{10}\right)=\sin\frac{\pi}{10}$

$\sin\frac{13\pi}{10}=\sin\left(\pi+\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}$ and $\sin\frac{17\pi}{10}=\sin\left(2\pi-\frac{3\pi}{10}\right)=-\sin\frac{3\pi}{10}$

So, $\sin\frac{\pi}{10},-\sin\frac{3\pi}{10}$ are the roots of $(2)$

As $0<\frac{\pi}{10}<\frac\pi2, \sin\frac{\pi}{10}>0\implies \sin\frac{\pi}{10}=\frac{-1+\sqrt5}4$

As $0<\frac{3\pi}{10}<\frac\pi2, -\sin\frac{3\pi}{10}<0\implies -\sin\frac{3\pi}{10}=-\frac{1+\sqrt5}4$

Answer 3

For the first one:

$$\lim_{x\to 0}\frac{5^x-4^x}{x} = \frac{\mathrm{d}(5^x-4^x)}{\mathrm{d}x}(0) = \frac{\mathrm{d}5^x}{\mathrm{d}x}(0)-\frac{\mathrm{d}4^x}{\mathrm{d}x}(0) = \log_e\frac{5}{4}$$

For the second:

$$\sin 2\frac{\pi}{10} + \cos 3\frac{\pi}{10} = 0.$$

The golden-ratio triangle has angles $\frac{\pi}{5}$ and two $\frac{2\pi}{5}$, so $\cos\frac{\pi}{5} = \frac{1+\sqrt{5}}{4}$ and using $\cos 2\theta = 1-2\sin^2\theta$ we get

$$\sin\frac{\pi}{10} = \sqrt{\frac{1-\cos\frac{\pi}{5}}{2}} = \sqrt{\frac{3-\sqrt{5}}{8}} = \sqrt{\frac{5-2\sqrt{5}+1}{16}} = \frac{\sqrt{5}-1}{4} .$$

I hope this helps ;-)