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A tough limit

Multivariate Calculus
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Grigorios Kostakos
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A tough limit

#1

Post by Grigorios Kostakos » Thu Sep 14, 2017 11:19 am

Examine if the limit $$\displaystyle\mathop{\lim}\limits_{(x,y)\to(1,1)}{\frac{x^y-y\,(x-1)-1}{x^2+y^2-2\,(x+y-1)}}$$ exists and if does exists, calculate it.


NOTE: I've got an (almost) solution.
Grigorios Kostakos
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Grigorios Kostakos
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Re: A tough limit

#2

Post by Grigorios Kostakos » Tue Sep 19, 2017 7:48 pm

The function $f(x,y)=x^y\,,\; (x,y)\in{\mathbb{R}}^2\,,$ is at least twice continuously differentiable in a open disk with centre $(1,1)$ and its 2nd degree Taylor polynomial is
\begin{align*}
P_{2,f,(1,1)}(x,y)&=f(1,1)+({\rm{grad}}\,{f})(1,1)\cdot(x-1,y-1)+\frac{1}{2}\,(x-1,y-1)\,H_f(1,1)\,(x-1,y-1)^{\top}\\
&=1+(1,0)\cdot(x-1,y-1)+\frac{1}{2}\,(x-1,y-1)\,\begin{pmatrix}
0 & 1\\
1& 0
\end{pmatrix}\,(x-1,y-1)^{\top}\\
&=1+x-1+\frac{1}{2}\,2\,(x-1)(y-1)\\
&=y\,(x-1)+1\,,
\end{align*}
for which holds
\begin{align*}
\displaystyle\mathop{\lim}\limits_{(x,y)\to(1,1)}\frac{f(x,y)-P_{2,f,(1,1)}(x,y)}{\big\|(x-1,y-1)\big\|^2}=0 \quad \Rightarrow\\
\displaystyle\mathop{\lim}\limits_{(x,y)\to(1,1)}{\frac{x^y-y(x-1)-1}{x^2+y^2-2(x+y-1)}}=0\,.\end{align*}
Grigorios Kostakos
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