Prove the limit exist (Partial Differential)

In summary, you will need to differentiate sin x/x and cos x/x. You will need to use L'Hospital Rule and apply it to r=0. Lastly, you will need to find the limit for 0-.
  • #1
naspek
181
0

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?
 
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  • #2
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
[tex] \lim{r \to 0^+} \frac{sin(r^2)}{r} [/tex]

as another method, how about multiplying through by r/r then considering the limit ?
 
  • #3
hurmm.. ok..
so.. wut bout limit for 0-?
 
  • #4
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
 
  • #5
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..
 
  • #6
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...
 
  • #7
damn bro/sis
hahahhaa
r u one of mmu's cyberjaya student?
u got the answer for other question?
damn what a coincidence
 
  • #8
HallsofIvy said:
As you say, with [itex]u= x^2+ y^2+ z^2[/itex],
[tex]\frac{sin(x^2+y^2+ z^2)}{(x^2+ y^2+ z^2)^{1/2}}= \frac{sin(u^2)}{u}[/tex]
Multiply both numerator and denominator by u to write that as
[tex]u\frac{sin(u^2)}{u^2}[/tex]
and now use [itex]sin(\theta)/\theta[/itex].

how about the other questions?
 

Related to Prove the limit exist (Partial Differential)

1. What does it mean to prove the limit exists in a partial differential equation?

Proving the limit exists in a partial differential equation means that as the independent variables approach a certain value, the dependent variable also approaches a specific value. This indicates that the function is continuous at that point.

2. Why is it important to prove the limit exists in a partial differential equation?

Proving the limit exists in a partial differential equation is important because it helps us understand the behavior of a function at a specific point. It also allows us to determine if the function is well-defined and has a unique solution.

3. How do you prove the limit exists in a partial differential equation?

The most common method for proving the limit exists in a partial differential equation is to use the epsilon-delta definition of a limit. This involves showing that for any given small value of epsilon, there exists a corresponding value of delta that ensures the function stays within epsilon distance of the limit point.

4. What are some common techniques used to prove the limit exists in a partial differential equation?

Some common techniques used to prove the limit exists in a partial differential equation include using the squeeze theorem, using the definition of continuity, and using algebraic manipulation to simplify the equation and show that it satisfies the definition of a limit.

5. Are there any situations where the limit may not exist in a partial differential equation?

Yes, there are some situations where the limit may not exist in a partial differential equation. This can occur when there is a jump or discontinuity in the function, or when the limit approaches different values depending on the direction from which the independent variables are approaching the limit point.

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