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$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$

It is about the $(\to1)^{(\to\infty)}$ situation. Can we find its limit using the formula $\lim_{x\to\infty}(1+\frac 1x)^x=e$? If yes, then how?

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  • $\begingroup$ Hint: Use the Taylor expansion and stop it at the right place to get a workable formula. $\endgroup$ – Martigan Jun 3 '15 at 17:22
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Put $y=\dfrac{1}{x} \to \displaystyle \lim_{x\to \infty}\ln(f(x))=\displaystyle \lim_{y\to 0}\dfrac{\ln\left(\sin y+\cos y\right)}{y}=\displaystyle \lim_{y\to 0} \dfrac{\cos y - \sin y}{\sin y+ \cos y}=1\to \displaystyle \lim_{x \to \infty} f(x) = e$.

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  • $\begingroup$ Please explain how $\lim_{y\to0}\frac{\ln{\sin y +\cos y}}y = \lim_{y\to0}\frac{\cos y - \sin y}{\cos y + \sin y}$ ? $\endgroup$ – Arpan Biswas Jun 4 '15 at 1:39
  • $\begingroup$ Ok thanks ! I understood it. You used L Hospitals in that str $\endgroup$ – Arpan Biswas Jun 4 '15 at 1:46
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Set $1/x=h$ to get

$$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=\lim_{h\to 0^+}\left(\sin h+\cos h\right)^{\frac1h}$$

Now $(\sin h+\cos h)^2=1+2\sin h\cdot\cos h=1+\sin2h$

$$\lim_{h\to 0^+}\left(\sin h+\cos h\right)^{\frac1h}=\lim_{h\to 0^+}\left[\left(\sin h+\cos h\right)^2\right]^{\frac1{2h}}$$

$$=\left[\lim_{h\to 0^+}\left(1+\sin2h\right)^{\dfrac1{\sin2h}}\right]^{\lim_{h\to 0^+}\dfrac{\sin2h}{2h}}$$

Now for the inner limit, use $\lim_{ y\to0}(1+y)^{\dfrac1y}=\lim_{m\to\infty}\left(1+\dfrac1m\right)^m=e$

How about the limit in the exponent?

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$$\sin \frac1x + \cos \frac1x = 1 + \frac1x - \frac1{2x^2} + \cdots$$ therefore $$ \left( \sin \frac1x + \cos \frac1x \right)^x = \left( 1 + \frac1x+\cdots\right)^x \to e \text{ as } x \to \infty.$$

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  • $\begingroup$ I'm not the down voter. But I think it might be better to replace $\cdots$ with $o(\frac1x)$ in your answer. $\endgroup$ – Vim Jun 3 '15 at 17:37
  • $\begingroup$ @Vim, thanks for your support. the drawing in your profile, did you draw that? looks nice. $\endgroup$ – abel Jun 3 '15 at 17:39
  • $\begingroup$ Thanks. Yes that's the best sketch I've ever made :) $\endgroup$ – Vim Jun 3 '15 at 23:19
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$$ \lim_{x\to\infty}(\sin{\frac 1x}+\cos{\frac 1x})^x $$ Take $x=\dfrac{1}{t}$ , $$ \lim_{t\to 0}(\sin t+\cos t)^\frac{1}{t} $$ $$ \large \lim_{t\to 0}e^\frac{\ln(\sin t+\cos t)}{t} $$ $$ \large e^{\lim_{t\to 0}\frac{\ln(\sin t+\cos t)}{t}} $$ Then use L'Hôpital's rule.

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Hint:

$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x= \lim_{x\to\infty}\exp\left(x\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)\right)= $$ $$= \exp\lim_{x\to\infty}\left(x\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)\right)= $$ $$ =\exp\lim_{x\to\infty} \dfrac{\log\left(\sin{\frac 1x}+\cos{\frac 1x}\right)}{\frac 1x}= $$

Now use l'Hopital rule.

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$$\begin{array}{lll} \lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x&=&\lim_{h\to 0}\left(\sin{h}+\cos{h}\right)^{\frac{1}{h}}\\ &=&\lim_{h\to 0}(\cos h)^{\frac{1}{h}}\left(\tan{h}+1\right)^{\frac{1}{h}}\\ &=&\lim_{h\to 0}((1+\cos h-1)^{\frac{1}{\cos h -1}})^{\frac{\cos h -1}{h}}(\left(\tan{h}+1\right)^{\frac{1}{\tan h}})^\frac{\tan h}{h}\\ &=&e^{\lim_{h\to 0}{\frac{\cos h -1}{h}}}e^{\lim_{h\to 0}\frac{\sin h}{h}\cdot \lim_{h\to 0}\frac{1}{\cos h}}\\ &=&e^0e^{1\cdot 1}\\ &=&e\\ \end{array}$$

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Set $1/x=2h,$

$$F=\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x=\lim_{h\to0}(1+\sin2h+\cos2h-1)^{1/2h}$$

Now using $\sin2h=2\sin h\cos h,\cos2h=1-2\sin^2h$ $$\sin2h+\cos2h-1=2\sin h(\cos h-\sin h)$$

$$F=\left[\lim_{h\to0}\{1+2\sin h(\cos h-\sin h)\}^{1/2\sin h(\cos h-\sin h)}\right]^{\lim_{h\to0}\frac{\sin h(\cos h-\sin h)}h}=e^1$$

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