Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A function $f: \mathbb R^2 \to \mathbb R$ is defined as $$f(x,y) = \frac{x^3\sin(x+y) - y^4\ln(x^2+y^2)}{x^2+y^2}$$

if $(x,y) \neq (0,0)$ and $f(0,0)=0$.

How can I show it is continuous at $(0,0)$? Perhaps by using the Squeeze theorem?

share|cite|improve this question
it looks like my multivariable calculus exams hehe – A Ricko Maulidar Feb 3 '13 at 13:11
The title doesn't make sense. – Jonas Meyer Feb 4 '13 at 2:27

Note that $\displaystyle f(x,y) = \frac{x^3 \sin(x+y)}{x^2 + y^2} - \frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}$.

So it's enough to show that each of these tends to $0$.

Applying the squeeze theorem on the first

$ \displaystyle | \frac{x^3 \sin(x+y)}{x^2 + y^2}| \le \frac{|x|}{1 + \frac{y^2}{x^2}} \le |x|$.

Next $\displaystyle |\frac{y^4 \ln(x^2 + y^2)}{x^2 + y^2}| = |\frac{y^2 \ln(x^2 + y^2)}{\frac{x^2}{y^2} + 1} | \le |y^2 \ln(x^2 + y^2)|$.

Now set $x = r\cos(\theta)$ and $y = r \sin(\theta)$. Then the limit can be estimated by $|r^2 \ln(r^2)| \to 0$ as $r \to 0$ (say by L'Hopital) independent of $\theta$.

share|cite|improve this answer

you just need to prove $f(x,y)$ differentiable at $(0,0)$ then you can use theorem "if $f$ differentiable at $c$ then $f$ continuos at $c.$"

you can see the theorem at any analysis books

share|cite|improve this answer
How can I show it's continuous without showing it's differentiable first? – user60783 Feb 3 '13 at 13:12
i think it would be more difficult because you must use the definition ($\epsilon - \delta$) – A Ricko Maulidar Feb 3 '13 at 13:13
How do you propose to show that $f$ is differentiable at $(0,0)$? – Jonas Meyer Feb 4 '13 at 2:30
you just need prove $f_x(0,0)$ and $f_y(0,0)$ exist – A Ricko Maulidar Feb 4 '13 at 15:29
A Ricko Maulidar: That is incorrect. Existence of partial derivatives is not sufficient for continuity, let alone differentiability at a point. – Jonas Meyer Feb 4 '13 at 15:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.