How can I prove that $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{\mathrm{e}^{(a+x)(b+y)}-\mathrm{e}^{ab}-b\mathrm{e}^{ab}x-a\mathrm{e}^{ab}y}{\sqrt{x^2+y^2}}=0$
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Let $x = y$ for your limit – Varun Iyer Aug 04 '14 at 11:45
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That doesn't prove anything @VarunIyer, except for that single path. – Silynn Aug 04 '14 at 11:46
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But if you can prove it for more than one path, you can prove the limit exists and it equals 0 – Varun Iyer Aug 04 '14 at 11:47
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If by more than one path you mean every path, then yes. – Silynn Aug 04 '14 at 11:48
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not even if by all paths does the limit exist, http://math.stackexchange.com/q/829201/149912 – Rene Schipperus Aug 04 '14 at 11:50
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It is easy to prove that $f(x,y)=e^{(a+x)(b+y)}$ differentiable on $(0,0)$. So we have(by definition) $$f(x,y)-f(0,0)=f'_x(0,0)x+f'_y(0,0)y+o(\sqrt{x^2+y^2}).$$
pointer
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You should prove that the function is differentiable $\textbf{of course}$ directly only with the definition. – bigli Aug 04 '14 at 12:30
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I know it. Realy, when I like prove that function $\mathrm{e}^{xy}$ is differentiable in $(a,b)$ then I had to show that $\textbf{the limit}$ is $0$. Now, what do you say???? – bigli Aug 04 '14 at 12:38
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Please, you prove that $\mathrm{e}^{xy}$ is differentiable in all $(a,b)$ by the definition directly. – bigli Aug 04 '14 at 12:48
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It has continuous partial derivatives, so it is differentiable (Do you know that theorem? If not, you can try to prove it or find it in every calculus book that has a chapter about functions of many variables). – pointer Aug 04 '14 at 18:48
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Although I do believe that this should be considered an easy result from usual calculation rules, if you really want to do it by hand, use polar coordinates. It's usually easier to look at these problems in polar coordinates, because you go from 2 variables to just one.
Here: $$\lim_{(x,y)\to(0,0)}\dfrac{\mathrm{e}^{(a+x)(b+y)}-\mathrm{e}^{ab}-b\mathrm{e}^{ab}x-a\mathrm{e}^{ab}y}{\sqrt{x^2+y^2}}= \lim_{r\rightarrow 0} \frac{\exp(a+r\cos\theta)(b+r\sin\theta)-\exp(ab)(1+br\cos\theta + ar\sin \theta)}{r} = ...$$
Matt B.
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