5
$\begingroup$

Find this limit :

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

note $x=e^{\ln{x}}$ $$I=\exp\left(\lim_{x\to\infty}x\ln{\left(\sin{\frac{2}{x}}+\cos{\frac{1}{x}}\right)}\right)$$

and let $\frac{1}{x}=t$,then $$\lim_{t\to 0}\frac{\ln{(\sin{2t}+\cos{t})}}{t}=\lim_{t\to 0}\dfrac{2\cos{2t}-\sin{t}}{\sin{2t}+\cos{t}}=2$$ so $$I=e^2$$

My question: have other methods? Thank you

$\endgroup$
  • $\begingroup$ The Maple command $$ asympt((sin(2/x)+cos(1/x))^x, x)$$ outputs $$ {{\rm e}^{2}}-5/2\,{\frac {{{\rm e}^{2}}}{x}}+{\frac {131}{24}}\,{ \frac {{{\rm e}^{2}}}{{x}^{2}}}-{\frac {569}{48}}\,{\frac {{{\rm e}^{2 }}}{{x}^{3}}}+{\frac {29107}{1152}}\,{\frac {{{\rm e}^{2}}}{{x}^{4}}}+ O \left( {x}^{-5} \right) .$$ $\endgroup$ – user64494 Nov 22 '13 at 7:49
  • $\begingroup$ "other methods?" Yes, limited expansion of sine and cosine--but basically this is just avoiding the L'Hôpital step. (But $x=e^{\ln x}$ works only for $x\gt0$, while your limit is when $x\to\pm\infty$, fortunately you do not use it, actually). $\endgroup$ – Did Nov 22 '13 at 8:04
5
$\begingroup$

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

$$\implies(\sin2h+\cos h)^{\frac1h}=(\cos h)^{\frac1h}\cdot(1+2\sin h)^{\frac1h}$$

$$\text{Now, }\displaystyle\lim_{h\to0}(1+2\sin h)^{\frac1h}=\left(\lim_{h\to0}(1+2\sin h)^{\frac1{2\sin h}}\right)^{2\frac{\lim_{h\to0}\sin h}h}=e^2$$

using $\displaystyle\frac1u=n,\lim_{u\to0}\left(1+u\right)^{\frac1u}=\lim_{n\to\infty}\left(1+\frac1n\right)^n=e$

Again, $\displaystyle h=2v\implies\lim_{h\to0}(\cos h)^{\frac1h}=\lim_{v\to0}(\cos2v)^{\frac1{2v}}$

$$=\displaystyle \left(\lim_{v\to0}(1-2\sin^2v)^{-\frac1{2\sin^2v}}\right)^{(\lim_{v\to0}-\frac{2\sin^2v}{2v})}$$

Observe that the inner limit $=e$

and $\displaystyle\lim_{v\to0}-\frac{2\sin^2v}{2v}=-\left(\lim_{v\to0}\frac{\sin v}v\right)^2\cdot \lim_{v\to0} v=1\cdot 0$

$\endgroup$
5
$\begingroup$

Using Taylor expansion for $x\to\infty$, $\sin\left(\frac{2}{x}\right)\sim \frac{2}{x}$ and $\cos\left(\frac{1}{x}\right)\sim 1$ so that $$ \left[\sin\left(\frac{2}{x}\right)+\cos\left(\frac{1}{x}\right)\right]^x\sim \left(\frac{2}{x}+1\right)^x\to\operatorname{e}^2 $$

$\endgroup$
  • 1
    $\begingroup$ this is the way to go.. short and easy! $\endgroup$ – Ant Nov 22 '13 at 10:37

Your Answer

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