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I was wondering if there is another approach instead of using the $log (x + 1) \sim_{0} x$ or Taylor series to solve:

$$\lim_{(x,y) \to (0,0)} f(x,y)={\ln(xy + 1) \over x}.$$

Particularly, I'm wondering whether I can use polar coordinates and l'Hopital.

Something like this. First, we switch to polar coordinates:

$$\lim_{(r) \to (0)} f(r)={\ln(r^2\cos\phi \sin\phi + 1) \over r\cos \phi}.$$

Now we use l'Hopital:

\begin{align} &\lim_{(r) \to (0)} f(r)=\lim_{(r) \to (0)}{\ln(r^2\cos\phi \sin\phi + 1) \over r\cos \phi} \\ =&\lim_{(r) \to (0)}{{d\over dr}(\ln(r^2\cos\phi \sin\phi + 1) \over{d\over dr} (r\cos \phi)} \\ =&\lim_{(r) \to (0)}{2r\sin\phi \cos\phi \over {r^2\sin\phi \cos^2\phi + \cos\phi}} \\ =&\lim_{(r) \to (0)}{2r\sin\phi \over r^2\sin\phi \cos\phi +1}=0. \end{align}

As $$\lim_{(r) \to (0)} f(\phi)= {2r\sin\phi }= 0.$$

Does this make any sense at all or I can't use l'hopital in this case?

jjjx
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  • Why do you have $p$ in the denominator and $r$ in the numerator? Aren't they the same? I recommend $r$ or $\rho$. You can also put backslashes before the trig functions to get them in the correct font, like you did with $\ln$ – Ross Millikan Jun 04 '13 at 13:44
  • You have several errors: $r^2$ in numerator, $r$ in denominator. See http://math.stackexchange.com/questions/405884/lhopitals-rule-and-multivariable-limits/405947#405947 You can make this work if you can bound your result independent of $\theta$. – Ted Shifrin Jun 04 '13 at 13:51
  • I fixed some of the errors. Thank you for the heads up. – jjjx Jun 04 '13 at 14:28
  • @TedShifrin I fixed a couple of errors. Can you take a look now at my approach? – jjjx Jun 04 '13 at 14:32
  • This limit has been dealt with a few days ago already. See here: http://math.stackexchange.com/questions/404619/help-with-lim-x-y-to-0-0-lnxy-1-over-x/404640#404640 – Christian Blatter Jun 04 '13 at 15:11
  • @ChristianBlatter The question you are referring to was asked by me. Here I'm asking whether it is OK to solve it with polar coordinates and l'hopital. – jjjx Jun 04 '13 at 15:16

2 Answers2

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In L'Hospital rule it was assumed that the limit is of the form $\frac{0}{0}$ on the limit and in the latter case this isn't true. There is also another problem with calculating the limit radially. Consider the function $f(x,y) = \Big\{\begin{eqnarray} 1, & \ y = x^2 \\ 0 , & \ y \neq x^2 \end{eqnarray}$. This has the property $\lim_{r \rightarrow 0} f(r,\theta) = 0$ for every $\theta$, but $f$ has no limit at origo. Now recall the definition of limit at $x_0$. For every $\epsilon > 0$ there is $\delta > 0$ s.t. $|x-x_0| < \delta$ implies $|f(x)-a| < \epsilon$. That means that you have to calculate an estimate for finite neigborhoods of $x_0$ and then show that those estimates go to $0$. L'hospital rule isn't good for that purpose because it gives only the limit without knowledge of the estimates.

Juha-Matti Vihtanen
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  • I don't see why this function has the property that $f(r,\theta)\to 0$ for every $\theta$? Wouldn't it only go to zero for every $\theta$ such that $\sin(\theta)=r\cos^2(\theta)$? – michek Apr 13 '15 at 11:20
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A few issues:

Why are you using $r$ in some places and $p$ in others?

You are missing a factor of $r$ in your substitution of $x y$.

You should not be applying L'Hopital's rule a second time as your numerator and denominator do not both tend to zero or infinity as $r$ ($p$?) tends to zero. At this stage you can just substitute in $r$/$p$. (Although the expresssion will be different once you consider the above)

Otherwise I believe this is fine.

john
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  • I edited the question to fix some of the errors. Thank you. – jjjx Jun 04 '13 at 14:28
  • could you please have a look at my approach after the fixes I made? Thank you! – jjjx Jun 04 '13 at 14:33
  • No problem. That all seems fine now (except formally you should still be including the fact that you're taking a limit in each expression) – john Jun 04 '13 at 14:34
  • no problem at all. – john Jun 04 '13 at 14:43
  • No, this is not correct. Notice that when $\cos\phi = 0$, your fraction takes the form $0/0$, which is in fact the same with the original function in cartesian coordinates. The limit does not exist. Please read my essay at http://math.stackexchange.com/questions/405884/lhopitals-rule-and-multivariable-limits/405947#405947, which deals with the need to give a uniform bound independent of $\theta$ (your $\phi$) if you're going to go this route. – Ted Shifrin Jun 04 '13 at 17:48
  • which fraction are you referring to? Also, if you believe the limit does't exist, can you give an example of two paths through the origin with different function limits? – john Jun 05 '13 at 01:18