Question

Find the limit if it exists, or show that the limit does not exist.
lim (x^2+y^2)/(sqrt(x^2+y^2+1)-1)
(x,y)->(0,0)

I've tried approaching the limit from the line y = 0 and y = x. Then I thought I could apply the squeeze theorem. No luck. The book says the answer is 2

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)} \frac{ x^2+y^2 }{ \sqrt{ x^2+y^2+1}-1 }\]

Sorry about changing the answer so many times. Devil is in the details.

Answer 2

you end up with the result 0/0
use L'Hospital Rule
lim (x,y) to (0,0) 2x_2yy'/(1/2 sqrt(x^2 +y^2+1)^-1/2(2x+2yy')
lim (x,y) to (0,0) 2sqrt(x^2+y^2+1) = 1
been awhile check

Answer 3

2 not 1

Answer 4

Sorry for my ignorance, I haven't taken calculus in 6 years. From what you wrote I should differentiate the equation with respect to y?

Answer 5

yes using implicit differentiation

Answer 6

I'll get on that. Thank you.

Answer 7

try switching coordinates ,,, use polar coordinates

Answer 8

also try switching 1/x and 1/y and change limits to infinity and use this theorem
http://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits

Answer 9

Thanks. I'm currently trying to work through KenLJW's suggestion and make sure I'm doing everything correctly. After that I'll look up polar coordinates again and try that approach.

Answer 10

it's a double sequence, you can'f differentiate 'y' w.r.t. 'x', they are independent coordinates ... dy/dx is zero naturally.

Answer 11

The approaches we've covered to this point are l'hopital's, squeeze, and approaching from different lines. So far I'm working with l'hopital to try and get the correct answer. This question is not assigned and it's not mentioned under the suggested problems. It's just in between suggested problems so I assumed I had the tools to complete it.

Answer 12

use approaching from different lines then, put y=mx, and see if you get same limit for different values of 'm'. if limit is same then the limit exists. If limit is different then it does not exist.

Answer 13

perhaps you can use L'hopital rule after that.

Answer 14

At this point my understanding is that there could always exist another line such that the limit is different because we can approach the point (0, 0) from an infinite number of possibilities.

Answer 15

I'll try y=mx with l'hopital's rule.

Answer 16

good luck!! I gotta sleep

Answer 17

Thanks!

Answer 18

So Ken, I'm unable to get the same values as you. From your post I read the intermediate step I should get is

\[\lim_{(x,y) \rightarrow (0,0)} \frac{2xyy'1/2 } { \sqrt(x^2+y^2+1)^{-\frac{1}{2}(2x+2yy')}}\], is this correct?

Answer 19

The way I did it the Numerator canceled?

Answer 20

N' = 2x+2yy'
D' = 1/2 sqrt(x^2+y^2+1)^-1/2 (2x+2yy')

Answer 21

I still don't get that. Using the product rule I get
a massive equation.

Answer 22

I think I'll bring this into office hours tomorrow. Thanks for your help guys.

Answer 23

It's not the product rule it's the chain rule in the denominator

Answer 24

Someone from class showed me how to solve it through substitution. Thanks for all your help!

\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} \frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1} \]
\[v=\sqrt{x^2+y^2+1}\]
\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} \frac{a^2-1}{a-1}\]
\[u=a-1\]
\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} \frac{\left(u+1\right)^2-1}{u}\]
\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} \frac{u+2}{u} = \lim_{\left(x,y\right) \rightarrow \left(0,0\right)} u+2\]
\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} u+2 = \lim_{\left(x,y\right) \rightarrow \left(0,0\right)} a -1 + 2\]
\[\lim_{\left(x,y\right) \rightarrow \left(0,0\right)} \sqrt{x^2+y^2+1}+1 = 2\]

Answer 25

typo in the above response. v = a

Answer 26

Same as my answer which was done direct using chain rule

Answer 27

that's same as using using polar coordinates http://www.wolframalpha.com/input/?i=limit+r-%3E0+r%5E2%2F%28sqrt%28r%5E2+%2B+1%29+-+1%29 although that substitution is clever than polar

Answer 28

Even with polar you still have to use L'Hospital and change rule if you don't graph

Answer 29

When you made substitution all you had to do was factor numerator then cancel giving direct answer

Answer 30

yes that it is, but I am not sure of method suggested. I've never seen using differentiation using on double limits

Answer 31

It's okay if x goes to 0 then y go to 0 is the same as y goes to 0 then s goes to 0

Answer 32

from either direction

Answer 33

f(x,y) = xy/(x^2 + y^4) limit (x,y) = (0,0) this case it's different.

Answer 34

f(x,y)=yx^3+xy^3
an apparent odd function? limit close by is not symmetric?

Answer 35

(x,y)->(0,0) your limit is zero .. i guess

Answer 36

I'm going on the assumption close to 0 must be symmetric near by.
why don't you graph it on your Mathematica program

Answer 37

the function is continuous at (x,y) = (0,0), naturally limit is zero.