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