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 August 13th, 2016, 12:36 PM #1 Newbie   Joined: Feb 2012 Posts: 19 Thanks: 0 A problem from Robert Osserman's book Dear all, Here is a problem from Osserman's book. Any one can help? Fix a point $\displaystyle (x_0, y_0)$ and let $\displaystyle d(x, y) = \sqrt{(x-x_0)^2+(y-y_0)^2}$. Show that: if $\displaystyle L(x, y) = Ax + By + C$, and if $\displaystyle \lim_{(x,y)->(x_0, y_0)}\frac{L(x,y)}{d(x,y)}=0$, then $\displaystyle A=B=C=0$. I can see $\displaystyle \lim_{(x,y)->(x_0, y_0)}L(x,y)=Ax_0+By_0+C=0$, but why must we have $\displaystyle A=B=C=0$? Thanks in advance.
 August 13th, 2016, 04:50 PM #2 Global Moderator   Joined: May 2007 Posts: 6,683 Thanks: 658 You need more to get the limit of L/d to be 0. All you have so far is 0/0, which is undetermined.
 August 13th, 2016, 06:34 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra As a first pass, I'd suggest that you see what happens when you approach $(x_0,y_0)$ along:$x=x_0$; $y=y_0$; $x-x_0=y-y_0$; and $x-x_0=-(y-y_0)$.
August 13th, 2016, 08:28 PM   #4
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Quote:
 Originally Posted by v8archie As a first pass, I'd suggest that you see what happens when you approach $(x_0,y_0)$ along:$x=x_0$; $y=y_0$; $x-x_0=y-y_0$; and $x-x_0=-(y-y_0)$.
From your "a" and "b", we know $\displaystyle Ax_0+By_0+C=0$.

From "c": $\displaystyle y=(x-x_0)+y_0$, we have $\displaystyle \lim_{x\rightarrow x_0}L(x, (x-x_0)+y_0)=\lim_{x\rightarrow x_0}Ax+B((x-x_0)+y_0)+C=\lim_{x\rightarrow x_0}(A+B)(x-x_0)=0$ (the next to the last equality follows from $\displaystyle Ax_0+By_0+C=0$).

From "d", we can similarly derive $\displaystyle \lim_{x\rightarrow x_0}L(x, -(x-x_0)+y_0)=\lim_{x\rightarrow x_0}Ax+B(-(x-x_0)+y_0)+C=\lim_{x\rightarrow x_0}(A-B)(x-x_0)=0$.

As long as $\displaystyle A+B$ and $\displaystyle A-B$ are some finite constants, the above limits are zero as $\displaystyle x\rightarrow x_0$. I still cannot see why A=B=0. Can you clarify? Thanks.

 August 13th, 2016, 09:24 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,599 Thanks: 2587 Math Focus: Mainly analysis and algebra In the case that $x_0=y_0=0$, for example, then $$\lim_{x \to 0} {L(x,0) \over d(x,0)} = \lim_{x \to 0} {ax+c \over \sqrt{x^2}} = {ax \over |x|} + {c \over |x|} \\ \lim_{y \to 0} {L(0,y) \over d(0,y)} = \lim_{y \to 0} {by+c \over \sqrt{y^2}} = {by \over |y|} + {c \over |y|}$$ For these limits to exist, we require that $c=0$ and for them to be equal (and they must be equal for $\lim \limits_{(x,y) \to (0,0)} {L(x,y) \over d(x,y)}$ to exist) we need that $a=-a =b=-b\implies a=b=0$. I suppose that it is possible to generalise this approach for other choices for $x_0$ and $y_0$. You may need to go to the $\delta-\epsilon$ definition of the limit of $L(x,y)$ to do this. Alternatively, a change of variables ($u=x-x_0$, $v=y-y_0$) might make things clearer (remember that $x_0$ and $y_0$ are constants in the limit expression). Note: I haven't solved the problem myself so I couldn't post the solution even if I wanted to. I'm just suggesting how I would look at the problem. Last edited by v8archie; August 13th, 2016 at 09:36 PM.
 August 14th, 2016, 02:02 PM #6 Global Moderator   Joined: May 2007 Posts: 6,683 Thanks: 658 Start from $\displaystyle Ax_0+By_0+C=0$. Because C is a constant $\displaystyle L(x,y)=A(x-x_0)+B(y-y_0)$ Since we are interested in the limits for $\displaystyle x->x_0\ and\ y->y_0$, consider the special case: $\displaystyle y=y_0$, then $\displaystyle \frac{L(x,y)}{d(x,y)} = A\ for\ x>x_0\ and \ = -A \ for\ x August 15th, 2016, 05:41 AM #7 Newbie Joined: Feb 2012 Posts: 19 Thanks: 0 Quote:  Originally Posted by mathman Start from$\displaystyle Ax_0+By_0+C=0$. Because C is a constant$\displaystyle L(x,y)=A(x-x_0)+B(y-y_0)$Since we are interested in the limits for$\displaystyle x->x_0\ and\ y->y_0$, consider the special case:$\displaystyle y=y_0$, then$\displaystyle \frac{L(x,y)}{d(x,y)} = A\ for\ x>x_0\ and \ = -A \ for\ x
Now I see, thank you very much!

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