I have to solve the limit $$\lim_{(x,y)\to(\pi,\pi)} \frac{sin x + sin y}{x+y-2\pi}$$ but I'm struggling to do it, always get 0/0 no matter what
#limit multivariable calculus
bitter falcon
tight relicBOT
Gaarco
thick sigil
Try seeing if the limit exists
The best trick is to do the limit following lines
Another thing that u can do is to change the coordinates
So just eliminate the 2pi and do the limit when (x,y) go to (0,0)
thick sigil
The best trick is to do the limit following lines
If two lines give u different answers the limit does not exist
clear perch
sin a + sin b use the identity and the fact that sin is 2pi peridoic
thick sigil
Been thinking about it and it definitely exists
Do you know the definition of a derivative in multivariable calculus?
Bcs that would help a lot
If not that's also ok
bitter falcon
Do you know the definition of a derivative in multivariable calculus?
yes i do
thick sigil
yes i do
great, so a good idea is to right sin(x) as an expansion in (0,0)
u just need the first term really
so u end up with
$\lim_{(x,y) \to (0,0)} \frac{x+y+R_1(x,y)+R_2(x,y)}{x+y}$
tight relicBOT
Arm
thick sigil
now you can separate this expression
and u only have to calculate $\lim_{(x,y) \to (0,0)} \frac{R_1(x,y)}{x+y}$
tight relicBOT
Arm
bitter falcon
Mhh, but my limit goes to (pi,pi), how can i use expansions here?
thick sigil
yh, ofc
bitter falcon
moreover the correct answer is -1
thick sigil
but since sin(x) is periodic u can simplify the equation
thick sigil
moreover the correct answer is -1
oh yh, bcs sin(x+pi) is negative -sin(x)
mb
but still
the same process
bitter falcon
Ok thanks
thick sigil
also, I think u can use the squeeze theorem
but this captures more the idea behind this limit
sharp onyx
Doesn't the identity $\lim\limits_{x\to 0}\frac{\sin(x)}{x} = 1$ suffice?
tight relicBOT
ℝafain
sharp onyx
\begin{align*}
&\lim\limits_{(x,y)\to(\pi,\pi)}\frac{\sin(x)+\sin(y)}{x+y-2\pi}\
=&-\lim\limits_{(x,y)\to(\pi,\pi)}\frac{\sin(x-\pi)+\sin(y-\pi)}{x+y-2\pi}\
=&-2\lim\limits_{(x,y)\to(\pi,\pi)}\frac{\sin\left(\frac{x+y-2\pi}{2}\right)\cos\left(\frac{x-y}{2}\right)}{x+y-2\pi}\
=&-\lim\limits_{z\to 0}\frac{\sin(z)}{z}\cdot\lim\limits_{(x,y)\to(\pi,\pi)}\cos\left(\frac{x-y}{2}\right)\
=&-1
\end{align*}
tight relicBOT
ℝafain
sharp onyx
with $z=\frac{x+y-2\pi}{2}$
tight relicBOT
ℝafain
bitter falcon
$sin(a-\pi) = (a-\pi)$ for $a \to \pi$ so
$$\lim_{(x,y) \to (\pi,\pi)} \frac{-sin(x-\pi)-sin(y-\pi)}{x+y-2\pi} =$$
$$\lim_{(x,y)\to(\pi,\pi)} \frac{-(x+y-2\pi)}{x+y-2\pi} = -1$$
tight relicBOT
Gaarco
sharp onyx
Same idea, I basically just avoided writing out the relatively vague first line
bitter falcon
Thanks everybody 🙂
thick sigil
$sin(a-\pi) = (a-\pi)$ for $a \to \pi$ so
$$\lim_{(x,y) \to (\pi,\pi)} \frac{-si...
yh, that's the idea behind the expansion proof (it's just a bit more rigorous)
thick sigil
**ℝafain**
we are using that, I just don't like defining the variable 'z'
and doing the limit when z goes to 0