Finding limit of function in 2 variables

TheSaxon

Homework Statement


lim as (x,y)-->(0,0) of sin(x^2+y^2)/(x^2+y^2)

Questions: Does limit exist and if so, what is it.

Homework Equations





The Attempt at a Solution



1. The professor instructed us to convert to polar coordinates to see if result depends on theta. If it does no limit.

Conversion:

lim as r-->0 of sin(r^2 cos^2(theta) + r^2 sin^2(theta)) / (r^2 cos^2(theta) + r^2 sin^2(theta))

This I think reduces to sin(r^2)/r^2. This implies that the limit exists, correct(since the answer doesn't depend on theta)? However, this doesn't help me with actually finding the limit, as plugging in 0 still yields indeterminate 0/0. What do i do?
 
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If you notice you can make a substitution!

say u=x^2+y^2

now your limit becomes:

lim_{(x,y)->(0,0)}\frac{\sin(x^2+y^2)}{x^2+y^2}
as (x,y)->(0,0) u->(0)
lim_{u->0} \frac{\sin u}{u}

and well you should know sin u/u by heart xD

also you could have done the substitution from sin(r^2)/r^2 actually it's almost the same thing since (x,y)->(0,0) (r,theta)->(0,0)

and theta is nowhere in there so you have lim r->0 sin(r^2)/r^2
 
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i guess I am just dumb, i didnt know that lim of sin u / u is 1. I am taking math after a 2 year hiatus :(
 
o well it's just a limit that sometimes comes up and when it does it's useful to know it. Also it's part of the proof of derivative of sin x.

also when you have a limit like sin(r^2)/r^2 it's a calculus 1 type limit so you can use L'Hospital's rule if you get an indeterminate form, but using it here you get 0.
 
dont u get 1? sin(r^2)/r^2 L'hopitaled becomes (cos(r^2)2r)/2r = cos(r^2) = 1?


Anyway, thanks for help
 
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o whops sorry i did the derivative 2x and got (2rsinr^2)/2r ><

but yes you could do it multiple ways but the substitution would be fastest since you can do it straight away.
 
You don't need to use L'Hopital. You know from Calc I that
\lim_{x\rightarrow 0}\frac{sin(x)}{x}= 1
Just think of r2 as x.
 
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