Mathematics Asked by Anne on January 5, 2021
I have to evaluate the following limit
$$ lim_{x to 0}{frac{sin( pi cos x)}{x sin x} }$$
My solution is:
$$ lim_{x to 0}{frac{sin( pi cos x)}{x sin x} }=lim_{x to 0}{frac{pi cos x}{x cdot x} }=+infty$$
But the correct result is $frac{pi}{2}$. I can’t understand where I’m making mistakes.
The error, as noticed in the comments, is that $picos x$ is not close to $0$.
Simpler to use asymptotic analysis: near $0$, we have $:cos x=1-dfrac{x^2}2+o(x^2)$, and $sin usim_0 u$, so begin{align} frac{sin(picos x)}{xsin x}&=frac{sinBigl(pi-cfrac{pi x^2}{2}+o(x^2)Bigr)}{x sin x} =frac{sinBigl(cfrac{pi x^2}{2}+o(x^2)Bigr)}{xsin x} \ &sim_0frac{cfrac{pi not{!x^2}}{2}+o(not{!x^2})}{not{!x^2}}=frac{pi}{2}+o(1) end{align}
Correct answer by Bernard on January 5, 2021
$$L=lim_{x to 0}{frac{sin( pi cos x)}{x sin x} }$$ $$L=lim_{x to 0}{frac{sin( pi cos x)}{x^2} }$$ By L'Hôpital's rule: $$L=-lim_{x to 0}{frac{cos( pi cos x)pi sin x}{2x} }$$ $$L=-lim_{x to 0}{frac{cos( pi cos x)pi }{2} }$$ $$implies L=dfrac {pi}2$$
Answered by Satyendra on January 5, 2021
Only using
Answered by enzotib on January 5, 2021
Use the identity $>picos x=pibigl(1-2sin^2{xover2}bigr)$ to obtain $${sin(picos x)over xsin x}={sinbigl(2pisin^2{xover2}bigr)over xcdot2sin{xover2}cos{xover2}}={piover2}cdot{sinbigl(2pisin^2{xover2}bigr)over2pisin^2{xover2}}cdot{tan{xover2}over{xover2}} .$$ On the right hand side the two last factors tend to $1$ when $xto0$. It follows that $$lim_{xto0}{sin(picos x)over xsin x}={piover2} .$$
Answered by Christian Blatter on January 5, 2021
By some elementary inequalities, near $x=0$ we have $$ cos(x)=1-x^2/2+O(x^4)$$In particular, near zero we have $$ frac{sin(picos(x))}{xsin(x)} = frac{sin(pi(1-x^2/2)+ O(x^4))}{xsin(x)} $$Use the sine-reflection and angle-addition: $$ frac{sin(pi(1-x^2/2)+O(x^4))}{xsin(x)} = frac{sin(pi/2 cdot x^2+O(x^4))}{xsin(x)} $$ $$ = frac{sin(pi/2cdot x^2)cos(O(x^4))}{xsin(x)}+ frac{cos(pi/2 cdot x^2)sin(O(x^4))}{xsin(x)} $$Using $lim_{thetato 0}frac{sin(theta)}{theta}=1$, the second term vanishes in the limit, and $cos(O(x^4))to 1$. So we have $$ lim_{xto 0}frac{sin(pi/2 cdot x^2)cdot 1}{xsin(x)} = lim_{xto 0}frac{sin(pi/2 cdot x^2)}{x^2}cdot frac{x}{sin(x)} =frac{pi}{2}cdot 1 =frac{pi}{2} $$
Answered by overrated on January 5, 2021
There are many ways in which you can calculate this. For example: L'Hospital's rule applied once will get you to
$$underset{xto 0}{text{lim}}-frac{pi (x sin ) (cos (pi (x cos )))}{x (cos x)+sin x}$$
(I leave the derivatives to you, as a warm up).
Now by product rule you can split the fraction as a product:
$$-pi left(underset{xto 0}{text{lim}}cos (pi (cos x))right) underset{xto 0}{text{lim}}frac{sin x}{x (cos x)+sin x}$$
Easily observing that the brackets limit is $-1$ which then makes $-pi$ to $pi$.
Now you can can multiply and divide by $x$ to get
$$pi lim_{xto 0} frac{color{red}{sin(x)}}{xcos(x) + sin(x)}cdot frac{x}{color{red}{x}}$$
The red zone goes to $1$. Collect $x$ over and above:
$$pi lim_{xto 0} frac{x}{xleft(cos(x) + frac{color{red}{sin(x)}}{color{red}{x}}right)}$$
Again a red zone that goes to $1$.
$x$ can be simplified. The cosine of $xto 0$ goes to $1$ leading to
$$frac{pi}{1+1} = frac{pi}{2}$$
Answered by Turing on January 5, 2021
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