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I have $lim(x,y)->(0,0)$ $\frac{e^{-x^2-y^2}-1}{x^2+y^2}$

so I did $\frac{e^{-(r^2cos(\theta)^2+r^2sin(\theta)^2)}-1}{r^2cos(\theta)^2+r^2sin(\theta)^2}$

$\frac{e^{r^2(-(cos(\theta)^2+sin(\theta)^2))}-1}{r^2(cos(\theta)^2+sin(\theta)^2)}$

$\frac{e^{-r^2}-1}{r^2}$

But got stuck after that. The answer is supposed to be -1, but I don't see how to get that from here. Did I do it correctly up until this point even?

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  • $\begingroup$ There is something wrong with your solution. How did you go from step 3 to step 4. $\endgroup$ – user262291 Oct 8 '16 at 21:48
  • $\begingroup$ I also don't follow that step, but just pointing out that L'Hopitals rule would be helpful for the last step $\endgroup$ – rtpax Oct 8 '16 at 21:49
  • $\begingroup$ Remember that $(cos(x)^{2}sin(x)^{2})$ isn't equal to 1. $sin(x)^{2} + cos(x)^2 =1$ $\endgroup$ – user262291 Oct 8 '16 at 21:50
  • $\begingroup$ In fact, even without getting rid of every instance of $\theta$ we can use L'Hopitals rule with two partial derivatives with respect to $r$ then plugging in 0 for $r$ and $\theta$ $\endgroup$ – rtpax Oct 8 '16 at 21:53
  • $\begingroup$ Ok, I make a mistake when typing out the problem, the cos and sin shouldn't have been multiplied by each other. Post fixed. And thanks rtpax. $\endgroup$ – windy401 Oct 8 '16 at 21:55
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That $\lim_{u\to 0} (e^{-u}-1)/u = -1$ is just the definition of the derivative of $e^{-u}$ at $0.$ As $(x,y)\to (0,0),$ $x^2 + y^2 \to 0.$ Therefore

$$\lim_{(x,y) \to (0,0)}\frac{e^{-(x^2+y^2)}-1}{x^2 + y^2} = -1.$$

No need for polar coordinates.

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  • $\begingroup$ Oh. Well the question said to use polar coords. Hmm alright then. Thanks. $\endgroup$ – windy401 Oct 8 '16 at 23:11

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