2

Please consider this limit question

$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$

How should I solve this? I have no idea where to start please help.

Josh Corner
  • 59
  • $x\to \infty $ so $\frac{1}{x} \to 0$, here you can use $\lim_{x\to 0} \sin(x)/x = 1$ – jonsno May 03 '18 at 07:10
  • 4
    maybe it could be better use the term difficult instead of complex to avoid initial misunderstanding reading the title, even if it is not difficult at all by standard limits – user May 03 '18 at 07:40

5 Answers5

7

HINT

Note that

$$\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}=\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{\frac{a}2\cdot \frac{\sin\left(\frac{a}{2x}\right)}{\frac{a}{2x}}}$$

user
  • 154,566
  • personally I find the below hint more intuitive, skipped too much. – Josh Corner May 03 '18 at 07:28
  • @JoshCorner It is indeed just a hint, note that the "difficult" part is only at the denominator and we can handle that by standard limit, how it is pointed out from user284331, since $a(x+1)/2x \to a/2$ the upper part is trivial. – user May 03 '18 at 07:33
  • @JoshCorner In any case it is not bad receive different hint by different answers for you I guess! Most important that now it is clear to you. – user May 03 '18 at 07:37
3

Make life easier using $x=\frac 1 y$.

So the problem reduces to $$\lim_{y\rightarrow 0}\left(\frac{a\, y}{\sin\left(\frac{a y}{2}\right)} \sin \left(\frac{a}{2}\right) \sin \left(\frac{a}{2} (y+1)\right) \right)$$ and the first term looks quite familiar.

Claude Leibovici
  • 260,315
2

A hint is given in the comment box, another hint is that $\lim_{x\rightarrow\infty}\dfrac{a(x+1)}{2x}=\dfrac{a}{2}$, then $\lim_{x\rightarrow\infty}\sin\left(\dfrac{a(x+1)}{2x}\right)=\sin\left(\lim_{x\rightarrow\infty}\dfrac{a(x+1)}{2x}\right)=\sin\left(\dfrac{a}{2}\right)$.

The fact that one can swipe the limit and the $\sin$ function needs some justification, essentially it is about the continuity of $\sin$ at any point, in this case, it is at the point $a/2$:

$|\sin u-\sin(a/2)|<\epsilon$ for all $|u-a/2|<\delta$, now choose a large $M>0$ such that $\left|\dfrac{a(x+1)}{2x}-\dfrac{a}{2}\right|<\delta$ for all $x\geq M$, then for such an $x$, one has $\left|\sin\left(\dfrac{a(x+1)}{2x}\right)-\sin\left(\dfrac{a}{2}\right)\right|<\epsilon$.

user284331
  • 55,591
1

Trivial with equivalents:

as $x\to\infty$,

  • $\dfrac{a(x+1)}{2x}\to \dfrac a2$, so $\sin\dfrac{a(x+1)}{2x}\sim_\infty \sin \dfrac a2$,
  • $\sin\dfrac{a}{2x}\sim_\infty\dfrac{a}{2x} $, so that $$\frac{a\sin \dfrac{a(x+1)}{2x}\,\sin\dfrac{a}{2}}{x\cdot \sin \dfrac{a}{2x}}\sim_\infty\frac{a\sin^2\dfrac a2}{x\,\dfrac a{2x}}=2\sin^2\dfrac a2=1-\cos a.$$
Bernard
  • 175,478
0

$$\lim_{x\rightarrow\infty}\frac{a\color{#f00}{\sin(\frac{a(x+1)}{2x})}\cdot \sin(\frac{a}{2})}{\color{#00ff}{x\cdot \sin(\frac{a}{2x})}}\tag{1}$$


$\color{#f00}{\sin\left(\frac{a(x+1)}{2x}\right)} = \sin\left(\frac{a}{2}+\frac{a}{2x}\right);\quad$ so $x \to \infty \implies \sin\left(\frac{a}{2}+0\right) = \sin\left(\frac{a}{2}\right)$


L'Hopital:
$\color{#00ff}{x\sin\left(\frac{a}{2x}\right)} = \left(\frac{\sin \left(\frac{a}{2x}\right)}{\frac{1}{x}}\right) =\left(\frac{-\frac{a\cos \left(\frac{a}{2x}\right)}{2x^2}}{-\frac{1}{x^2}}\right)=\left(\frac{a}{2}\cos \left(\frac{a}{2x}\right)\right);\quad$ so $x \to \infty \implies\frac{a}{2}\cos(0) = \frac{a}{2}$

$$\frac{a\color{#f00}{\sin\left(\frac{a}{2}\right)}\cdot \sin(\frac{a}{2})}{\color{#00ff}{\frac{a}{2}}}\tag{2}$$ $$2\sin ^2\left(\frac{a}{2}\right)\tag{3}$$

$$1-\cos\left(a\right)\tag{3}$$

Darío A. Gutiérrez
  • 3,863
  • 1
    In my opinion l'Hopital should be avoided when we can simply refer to standard limits. – user May 03 '18 at 08:22
  • 1
    @gimusi Unless you can show a discontinuity or other feature at which L'Hopital does not apply, why not use a valuable theorem? – Carl Witthoft May 03 '18 at 14:41
  • @CarlWitthoft It of course a valuable theorem which is needed also to proof Taylor's theorem but I do not consider that a good tool to solve limit since it acts as a black box machine and it doesn't give any enlightments about the role that each term plays in the to determine the limit. For that, I always suggest when possible to use standard limits for simple limits and Taylor's expansion for limits of higher order. In some cases l'Hopital can be very useful and effective, as for example for limit which involve integral functions. – user May 03 '18 at 14:48
  • 1
    @CarlWitthoft Of course it is a my own personal opinion and suggestion based on my small experience and it is mainly motivated by an educational issue. Standard limits and Taylor's expansion give a better feeling of the role played by each term in the expression. That's all, nothing personal with l'Hopital! – user May 03 '18 at 14:53