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I'm trying to show continuity of the function $$\frac {\ln(1+x^2+y^2)}{x^2+y^2}$$ for $(x,y)\neq 0$, $$f(x,y)=1$$ for $(x,y)= 0$, on $\mathbb{R}^2$ But I am not able to. The numerator is stuck for me as I don't know how to get around the log function. How can I simplify this expression or better get rid of the log? Sorry for the sloppy edits

Jacob
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  • If you want to establish the continuity everywhere, you need to define the function at $(0,0)$. If this is an exercise, and the problem was correctly posed, the function was probably defined to be $1$ at $(0,0)$. – André Nicolas Aug 16 '15 at 18:58
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    With the change, we do have continuity at $(0,0)$. All you need to do is show that $\lim_{t\to 0}\frac{\ln(1+t)}{t}=1$. This can be done in various ways, L'Hospital's Rule and the definition of derivative are two of them. – André Nicolas Aug 16 '15 at 19:06
  • Yes, it's so simple that way actually. Thanks – Jacob Aug 16 '15 at 19:10
  • You are welcome. – André Nicolas Aug 16 '15 at 19:15

4 Answers4

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HINT:

The function has a removable discontinuity at $(0,0)$ since $\lim_{(x,y)\to(0,0)}\frac{\log (1+x^2+y^2)}{x^2+y^2}=\lim_{z\to 0}\frac{\log (1+z)}{z}=1$.

For all $x^2+y^2\ne 0$, view $x^2+y^2$ as a single variable and show continuity of $f(z)=\frac{\log (1+z)}{z}$ for $z\ne 0$

Mark Viola
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Hint: $$ \lim_{(x,y)\rightarrow (0,0)}\frac {\ln(1+x^2+y^2)}{x^2+y^2}=1 $$

Emilio Novati
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Hint: $$\lim _{ x\rightarrow 0\\ y\rightarrow 0 }{ \frac { \ln { \left( 1+{ x }^{ 2 }+{ y }^{ 2 } \right) } }{ { x }^{ 2 }+{ y }^{ 2 } } } =\lim _{ x\rightarrow 0\\ y\rightarrow 0 }{ \ln { \left( 1+{ x }^{ 2 }+{ y }^{ 2 } \right) ^{ \frac { 1 }{ { x }^{ 2 }+{ y }^{ 2 } } } } } =\ln { e=1 } $$

haqnatural
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  • What? Is this true or are you kidding? I only ask since you're using a log property I have never seen and it almost seems like wizardry to me. Is there a rule for what you do with the exponent? – Jacob Aug 17 '15 at 19:18
  • is this wizard for you $\ln { { a }^{ b }=b\ln { a } } $ :)? you can read and learn from here https://en.wikipedia.org/wiki/Logarithm – haqnatural Aug 17 '15 at 19:27
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The function $f(x,y)=\dfrac{\ln(1+x^2+y^2)}{x^2+y^2}$ is defined and continuous on $\mathbf R^2\smallsetminus\{(0,0)\}$, as a composition and quotient of continous functions (polynomials and logarithms are continuous functions).

We can define a continuous continuation of $f$ at $(0,0)$ because $f$ has a limit at $(0,0)$. To see this, use polar coordinates: set $x=r\cos \theta,\ y=r\sin\theta$. Then for $(x,y)\neq (0,0)$, $$f(x,y)=\frac{\ln(1+r^2)}{r^2}\xrightarrow[r\to0]{}1. $$ Thus we obtain a continuous fonction on $\mathbf R^2$ if we set $$f(x)=\begin{cases}\dfrac{\ln(1+x^2+y^2)}{x^2+y^2}&\text{if}\enspace (x,y) \neq (0,0),\\1&\text{if}\enspace (x,y) = (0,0).\end{cases}$$

Bernard
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