Prove the limit exist (Partial Differential)

[naspek]

Homework Statement

Hi everyone this is my first time here - I'm completely confused not so I thought I may as well ask complete stangers for help!

the question is..
Determine whether the following limit exists. If so, find its value.

lim [sin(x^2+y^2+z^2)] / (x^2+y^2+z^2)^1/2
x,y,z-->(0,0,0)

Homework Equations

sin x/x = 1

The Attempt at a Solution

ok this is my idea..
if i let t=x^2+y^2+z^2, the limit should be like..

lim (sin t) / (t^1/2)

apply L'Hospital rule,

(cos t) / (t^-1/2) = t^1/2 cos t

am i right?

t=0.. so...

(cos t) / (t^-1/2) = t^1/2 cos t
= 0^1/2 cos 0
= 0
ok.. how am i going to proceed?

 

[lanedance]

first be a little careful, if you let t=x^2+y^2+z^2, then you are only looking at the limit form t->0+

that said i think your approach is valid, though you have missed a factor of -2 in your L'Hop differentiation, but you still get the limit is zero...

to see what is going on, if you let r = (x^2+y^2+z^2)^(1/2), you are essentially changing to spherical co-ordinates, where the limit is independent of theta or phi
\lim{r \to 0^+} \frac{sin(r^2)}{r}

as another method, how about multiplying through by r/r then considering the limit ?

 

[naspek]

hurmm.. ok..
so.. wut bout limit for 0-?

 

[lanedance]

i was just trying to clarify for your variable change you only consider t->0+, so you don't have to worry about 0-. Its not hugely important here, just something to be aware of

 

[naspek]

ic...
ok... let me refresh back my calculation...

lim ...[sin(x^2+y^2+z^2)] / (x^2+y^2+z^2)^1/2
x,y,z-->(0,0,0)

apply L'hospital rule..
so.. i need to differentiate nominator, f(x) and denominator, g(x)
let r = (x^2+y^2+z^2)^(1/2)
so.. i shall have sin r^2/r

then, (2r cos r^2)/1

substitute limit r=0
i'll get 2(0) cos 0
then the answer is 0..
so.. the limit is exist and the value is 0..

can u please help to confirm my answer lanedance..

 

[learnerx]

can we apply a conjugate into this question?

the answer will be:
you'll get the sin x/x = 1 thingy.

then your answer will be left with:
lim....(x^2 + y^2 + z^2)^(1/2)
(x,y,z)->(0,0,0)

and yada, yada, yada...

 

[iamyes]

damn bro/sis
hahahhaa
r u one of mmu's cyberjaya student?
u got the answer for other question?
damn what a coincidence

 

[iamyes]

As you say, with u= x^2+ y^2+ z^2,
\frac{sin(x^2+y^2+ z^2)}{(x^2+ y^2+ z^2)^{1/2}}= \frac{sin(u^2)}{u}
Multiply both numerator and denominator by u to write that as
u\frac{sin(u^2)}{u^2}
and now use sin(\theta)/\theta.

how about the other questions?