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August 13th, 2016, 11:36 AM  #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 $\displaystyle d(x, y) = \sqrt{(xx_0)^2+(yy_0)^2}$. 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, 03:50 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616 
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, 05:34 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 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:

August 13th, 2016, 07:28 PM  #4  
Newbie Joined: Feb 2012 Posts: 19 Thanks: 0  Quote:
From "c": $\displaystyle y=(xx_0)+y_0$, we have $\displaystyle \lim_{x\rightarrow x_0}L(x, (xx_0)+y_0)=\lim_{x\rightarrow x_0}Ax+B((xx_0)+y_0)+C=\lim_{x\rightarrow x_0}(A+B)(xx_0)=0$ (the next to the last equality follows from $\displaystyle Ax_0+By_0+C=0$). From "d", we can similarly derive $\displaystyle \lim_{x\rightarrow x_0}L(x, (xx_0)+y_0)=\lim_{x\rightarrow x_0}Ax+B((xx_0)+y_0)+C=\lim_{x\rightarrow x_0}(AB)(xx_0)=0$. As long as $\displaystyle A+B$ and $\displaystyle AB$ 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, 08:24 PM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,445 Thanks: 2499 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 ($u=xx_0$, $v=yy_0$) 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. 
August 14th, 2016, 01:02 PM  #6 
Global Moderator Joined: May 2007 Posts: 6,607 Thanks: 616 
Start from $\displaystyle Ax_0+By_0+C=0$. Because C is a constant $\displaystyle L(x,y)=A(xx_0)+B(yy_0)$ 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 $\displaystyle x=x_0$. 
August 15th, 2016, 04:41 AM  #7  
Newbie Joined: Feb 2012 Posts: 19 Thanks: 0  Quote:
 

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