2
$\begingroup$

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

$\endgroup$

4 Answers 4

2
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ $$=lim_{t \to 0}(1+(sin^2(t)-2sin^2(t/2)))^{\frac{(sin^2(t)-2sin^2(t/2))}{(sin^2(t)-2sin^2(t/2))}\cdot\frac{1}{t^2}} = (e^ \alpha) $$ because $$=lim_{\left(f(t) \to 0\right)}\left(1 + f(t)\right)^{\frac{1}{f(t)}} = e$$ correct? $\endgroup$
    – Jneven
    Dec 25, 2018 at 13:25
  • 1
    $\begingroup$ 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 $\endgroup$ Dec 25, 2018 at 16:13
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ This one uses Taylor which is, unfortunately a "forbidden" technique (I'm referring to the use of $O(u)$) $\endgroup$
    – Jneven
    Dec 25, 2018 at 13:29
1
$\begingroup$

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

$\endgroup$
1
  • $\begingroup$ This one uses Taylor which is, unfortunately a "forbidden" technique $\endgroup$
    – Jneven
    Dec 25, 2018 at 13:17
1
$\begingroup$

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

The inner limit converges to $e,$ right?

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.