Prove: $\lim\limits_{x\to\infty}\left(\sin^2\left(\frac{1}{x}\right)+\cos\frac1x\right)^{x^{2}}=\sqrt{e}$

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 $$\lim\limits_{x \to 0} \sin^2(x) = 0$$.

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

• $$=lim_{t \to 0}(1+(sin^2(t)-2sin^2(t/2)))^{\frac{(sin^2(t)-2sin^2(t/2))}{(sin^2(t)-2sin^2(t/2))}\cdot\frac{1}{t^2}} = (e^ \alpha)$$ because $$=lim_{\left(f(t) \to 0\right)}\left(1 + f(t)\right)^{\frac{1}{f(t)}} = e$$ correct? – Jneven Dec 25 '18 at 13:25
• 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 – Ankit Kumar Dec 25 '18 at 16:13

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

• This one uses Taylor which is, unfortunately a "forbidden" technique (I'm referring to the use of $O(u)$) – Jneven Dec 25 '18 at 13:29

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

• This one uses Taylor which is, unfortunately a "forbidden" technique – Jneven Dec 25 '18 at 13:17

$$\lim_{x\to\infty}\left(\sin^2\dfrac1x+\cos\dfrac1x\right)^{x^2}=\left(\lim_{x\to\infty}\left(1+\sin^2\dfrac1x+\cos\dfrac1x-1\right)^{1/(\sin^2\frac1x+\cos\frac1x-1)}\right)^{\lim_{x\to\infty}x^2(\sin^2\frac1x+\cos\frac1x-1)}$$

The inner limit converges to $$e,$$ right?

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