Question

Study the existence of the limit

Answer 1

\[\lim_{(x,y) \rightarrow (0,0)}\frac{ \ln(1+4x^2+9y^2) }{ 6xy }\]

Answer 2

@ganeshie8 @ParthKohli @Abhisar @cwrw238 @PRAETORIAN.10 @AriPotta @radar @eliassaab @Kainui @wio

Answer 3

One option is to use polar coords

Answer 4

I dunno how much that would help here.

Answer 5

It's an indeterminate form

Answer 6

` in this case the function is not continuous at the point in question and so we can’t just plug in the point. So, since the function is not continuous at the point there is at least a chance that the limit doesn’t exist. If we could find two different paths to approach the point that gave different values for the limit then we would know that the limit didn’t exist. Two of the more common paths to check are the x and y-axis so let’s try those.`

http://tutorial.math.lamar.edu/Classes/CalcIII/Limits.aspx

Lets try and see if the limit is DNE by finding two paths that approach to different values

Answer 7

y=1/x should do the trick

Answer 8

how ?

Answer 9

im thinking of straight lines through origin y = mx

Answer 10

Along y = mx, the limit becomes :
\[\lim\limits_{x\to 0} \frac{\ln (1+4x^2+ 9(mx)^2)}{6x(mx)} \stackrel{LH..}{=}\frac{9m^2+4}{6m}\]

Answer 11

Clearly the limit depends on "m",
you get different values by varying m. so the limit is DNE.

Answer 12

What if we construct this limit ln(1+x)/x ?

Answer 13

How did you evaluate the limit to that?

Answer 14

you want the limit to evaluate to same value from for ALL the paths that go approach (0, 0)

finding two paths that converge to same value will not do

Answer 15

wio, apply l'hopital then directly plugin x=0

Answer 16

Did you apply it twice?

Answer 17

@ganeshie8 are you sure we can apply l'hopital on multivariable limits
?

Answer 18

only once

Answer 19

after plugging in y = mx, you're travelling along a specific path
so it became single variable

Answer 20

let me break the steps for you

Answer 21

@Alexander95 L'Hospital works if you parametrize it into a single variable.

Answer 22

\[\lim_{(x,y) \rightarrow (0,0)}\frac{ \ln(1+4x^2+9y^2) }{ 6xy }\]
If you differentiate with respect to \(x\), you get: \[
\lim_{(x,y)\to(0,0)}\frac{\frac{8x+18yy'}{1+4x^2+9y^2}}{6y+6xy'}
\]

Answer 23

\[\begin{align}\lim\limits_{x\to 0} \frac{\ln (1+4x^2+ 9(mx)^2)}{6x(mx)} &\stackrel{LH}{=}
\lim\limits_{x\to 0} ~\frac{1}{1+4x^2+9m^2x^2}\cdot \frac{8x + 18m^2x}{12mx}\\~\\
&=\lim\limits_{x\to 0} ~\frac{1}{1+4x^2+9m^2x^2}\cdot \frac{4+9m^2}{6m }\\~\\

\end{align}\]

plugin x = 0

Answer 24

so for m<1 we got something negative and for m>1 we got something positive therefore the limit DNE @ganeshie8 Am I right ?

Answer 25

yeah if you take even two paths and one path has a different limit from another path then the limit doesn't exist

Answer 26

Suppose you have checked all linear paths and they converge to the same limit.
Would you still need to check other kinds of paths? I wonder.

Answer 27

that's scary

Answer 28

this is not useful for finding the limit
only works for proving DNE..

Answer 29

Would the path \(y=f(x)\) be equivalent to its tangent line path ?
\(L=f'(x_0)(x-x_0)+f(x_0)\). I am guessing not.

Answer 30

One thing I was doing with this problem:\[
(x,y)=\left(\frac r2\cos\theta,\frac r3\sin\theta\right)
\]But then I got stuck: \[
\lim_{r\to0}\frac{6\ln(1+r^2)}{r^2\cos\theta\sin\theta }
\]I should have used l'Hospital here to get: \[
\frac{6}{\sin\theta\cos\theta}
\]Which shows that \(\theta\), the angle you ride in on, determines the limit.

Answer 31

Take the paths x=y and x=-y.
The limit on x=y is 13/6
The limit on x=-y is -13/6

Answer 32

@ganeshie8

Answer 33

Ahh nice xD this looks much simpler and just what is necessary ! thanks @eliassaab

Answer 34

YW @ganeshie8