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August 13th, 2016, 11:36 AM   #1
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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.
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August 13th, 2016, 03:50 PM   #2
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You need more to get the limit of L/d to be 0. All you have so far is 0/0, which is undetermined.
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August 13th, 2016, 05:34 PM   #3
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As a first pass, I'd suggest that you see what happens when you approach $(x_0,y_0)$ along:
  1. $x=x_0$;
  2. $y=y_0$;
  3. $x-x_0=y-y_0$; and
  4. $x-x_0=-(y-y_0)$.
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August 13th, 2016, 07:28 PM   #4
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Quote:
Originally Posted by v8archie View Post
As a first pass, I'd suggest that you see what happens when you approach $(x_0,y_0)$ along:
  1. $x=x_0$;
  2. $y=y_0$;
  3. $x-x_0=y-y_0$; and
  4. $x-x_0=-(y-y_0)$.
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.
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August 13th, 2016, 08:24 PM   #5
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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 08:36 PM.
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August 14th, 2016, 01:02 PM   #6
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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 $\displaystyle \frac{L(x,y)}{d(x,y)} = A\ for\ x>x_0\ and \ = -A \ for\ x<x_0$. For the limit to be 0, A=0. Similarly one can show B=0 using the special case $\displaystyle x=x_0$.
Thanks from loveinla
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August 15th, 2016, 04:41 AM   #7
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Quote:
Originally Posted by mathman View Post
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 $\displaystyle \frac{L(x,y)}{d(x,y)} = A\ for\ x>x_0\ and \ = -A \ for\ x<x_0$. For the limit to be 0, A=0. Similarly one can show B=0 using the special case $\displaystyle x=x_0$.
Now I see, thank you very much!
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