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Prove - Without L'Hopital or Taylor: $$\lim\limits_{x\to\infty}\left(\sin^2\left(\frac{1}{x}\right)+\cos\frac1x\right)^{x^{2}}=\sqrt{e}$$

My Attempt:

$t \triangleq \frac{1}{x} $ such that:

$$\lim_{x \to \infty} \left(\sin^2 \left(\frac{1}{x}\right)+\cos\frac{1}{x}\right)^{x^{2}}= \lim_{t \to 0} \left(\sin^2(t)+ \cos(t) \right)^ {\frac{1}{t^2}}$$

$$ = \lim_{t \to 0} \ e^{ \left(\sin^2(t)+ \cos(t) \right)^{\frac{1}{t^2}}} = e ^{ \lim_{t \to 0} \ln \left(\sin^2(t)+ \cos(t) \right)^{\frac{1}{t^2}}} = e^ { \lim_{t \to 0} \ln \left(\sin^2(t)+ \cos(t) \right)\cdot \frac{1}{t^2}} $$

At this point - how do I imply some algebraic "trick" to prove that this limit is equal to a limit of the form: $\lim\limits_ { x \to 0} \left(1 +x\right)^{1/x^2}$, because $\lim\limits_{x \to 0} \sin^2(x) = 0$.

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$$L= lim_{t \to 0} \left(\sin^2(t)+ \cos(t) \right)^ {\frac{1}{t^2}}$$ $$=lim_{t \to 0} \left(\sin^2(t)+ 1-2sin^2(t/2) \right)^ {\frac{1}{t^2}}$$ $$=lim_{t \to 0}(1+(sin^2(t)-2sin^2(t/2)))^{\frac{(sin^2(t)-2sin^2(t/2))}{(sin^2(t)-2sin^2(t/2))}\cdot\frac{1}{t^2}}$$ $$=(e^{\alpha})$$ where $$\alpha=lim_{t\to 0}\frac{(sin^2(t)-2sin^2(t/2))}{t^2}=1/2.$$ So, $$L=\sqrt e$$

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  • $\begingroup$ $$=lim_{t \to 0}(1+(sin^2(t)-2sin^2(t/2)))^{\frac{(sin^2(t)-2sin^2(t/2))}{(sin^2(t)-2sin^2(t/2))}\cdot\frac{1}{t^2}} = (e^ \alpha) $$ because $$=lim_{\left(f(t) \to 0\right)}\left(1 + f(t)\right)^{\frac{1}{f(t)}} = e$$ correct? $\endgroup$ – Jneven Dec 25 '18 at 13:25
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    $\begingroup$ Yes! I thought you might already know this result, and used it directly without explicitly writing. In fact, this is one of the definitions of $e$ too $\endgroup$ – Ankit Kumar Dec 25 '18 at 16:13
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$$ \left(\sin^2(t)+ \cos(t) \right)^ {\frac{1}{t^2}}=e^{\frac{1}{t^2}\ln(\sin^2(t)+\cos(t))} $$ and the exponent $$ \frac{1}{t^2}\ln(\sin^2(t)+\cos(t))\sim\frac{1}{t^2}\ln\left(t^2+1-\frac{t^2}{2}+o(t^2)\right)\sim\frac{1}{t^2}\frac{t^2}{2}\ , $$ using $\sin(t)\sim t$, $\cos(t)\sim 1-t^2/2$ and $\ln(1+t)\sim t$ as $t\to 0$.

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  • $\begingroup$ This one uses Taylor which is, unfortunately a "forbidden" technique (I'm referring to the use of $O(u)$) $\endgroup$ – Jneven Dec 25 '18 at 13:29
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Hint: write $u={1\over u}$, $sin(u)=u+O(u^2), cos(u)=1-u^2/2+O(u^3)$ implies that $sin(u)^2+cos(u)=u^2/2+O(u^3)$,

$lim_{u\rightarrow 0}(sin(u)^2+cos(u))^{1\over u^2}=lim_{u\rightarrow 0}e^{{ln(sin(u)^2+cos(u))}\over u^2}=lim_{u\rightarrow 0}e^{{ln(1+u^2/2+O(u^3))}\over u^2}=e^{1\over 2}$

since $ln(1+u)=u+O(u)$

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  • $\begingroup$ This one uses Taylor which is, unfortunately a "forbidden" technique $\endgroup$ – Jneven Dec 25 '18 at 13:17
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$$\lim_{x\to\infty}\left(\sin^2\dfrac1x+\cos\dfrac1x\right)^{x^2}=\left(\lim_{x\to\infty}\left(1+\sin^2\dfrac1x+\cos\dfrac1x-1\right)^{1/(\sin^2\frac1x+\cos\frac1x-1)}\right)^{\lim_{x\to\infty}x^2(\sin^2\frac1x+\cos\frac1x-1)}$$

The inner limit converges to $e,$ right?

For the exponent set $1/x=h$ to find $$\lim_{h\to0^+}\dfrac{\sin^2h+\cos h-1}{h^2}=\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2-\lim_{h\to0^+}\dfrac1{1+\cos h}\left(\lim_{h\to0^+}\dfrac{\sin h}h\right)^2=1-\dfrac1{1+1}=?$$

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