# Math Help - prove the limit exist (Partial Differential)

1. ## prove the limit exist (Partial Differential)

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)

first, how can i prove that the limit is exist?

i'm thinking of to get the above limit to be something like
[sin (x)]/x = 1
or..
should i use quotient rule..?
hope that someone can guide me solving this question..

2. Originally Posted by nameck
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)

first, how can i prove that the limit is exist?

i'm thinking of to get the above limit to be something like
[sin (x)]/x = 1
or..
should i use quotient rule..?
hope that someone can guide me solving this question..
Yes, use spherical polar coords..

$
x = \rho \cos \theta \sin \phi,\;\;
y = \rho \sin \theta \sin \phi
,\;\;
z = \rho \cos \phi
$

so $x^2+y^2+z^2 = \rho^2$

3. i'm sorry..
i never learn that kind of equation..
got other ideas to solve it?

4. Originally Posted by nameck
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)

first, how can i prove that the limit is exist?

i'm thinking of to get the above limit to be something like
[sin (x)]/x = 1
or..
should i use quotient rule..?
hope that someone can guide me solving this question..
Is this $\lim_{(x,y,z)\to(0,0,0)}\frac{\sin\left(x^2+y^2+z^ 2\right)}{\left(x^2+y^2+z^2\right)^{\frac{1}{2}}}$ or $\lim_{(x,y,z)\to(0,0,0)}\left(\frac{\sin\left(x^2+ y^2+z^2\right)}{x^2+y^2+z^2}\right)^{\frac{1}{2}}$

5. Originally Posted by nameck
i'm sorry..
i never learn that kind of equation..
got other ideas to solve it?

$0 \le \sin \left(x^2+y^2+z^2\right) \le x^2+y^2+z^2$

so

$
0 \le \frac{\sin \left(x^2+y^2+z^2\right)}{\sqrt{x^2+y^2+z^2}} \le \sqrt{x^2+y^2+z^2}
$

Now letting $(x,y,z) \to (0,0,0)$ gives your answer.

6. 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?

7. so.. is my calculations above is correct?