- #1
Bashyboy
- 1,421
- 5
"As [itex](x,y) \rightarrow (0,0)[/itex], [itex]r \rightarrow[/itex]," is a fact that I am given, in order to solve a problem. I simply want to know if I properly understand why this fact is true.
I know that [itex]x = rcos \theta [/itex] and [itex]y=rsin \theta[/itex]. If I were to look at the individual limits, as [itex]x \rightarrow 0[/itex] and [itex]y \rightarrow 0[/itex], then I see each relation becomes [itex] 0 = rcos \theta [/itex] and [itex]0=rsin \theta[/itex]. Now, as x and y approach 0, there is no angle that [itex]\theta[/itex] approaches that make [itex]rcos \theta [/itex] and [itex]rsin \theta[/itex] simultaneously zero; hence, it must be that r approaches 0 as to make each expression zero simultaneously.
In summary, [itex]\lim_{(x,y) \rightarrow (0,0)}~(x+y) = lim_{r \rightarrow 0} ~ (rcos \theta + rsin \theta)[/itex].
Are my arguments correct?
I know that [itex]x = rcos \theta [/itex] and [itex]y=rsin \theta[/itex]. If I were to look at the individual limits, as [itex]x \rightarrow 0[/itex] and [itex]y \rightarrow 0[/itex], then I see each relation becomes [itex] 0 = rcos \theta [/itex] and [itex]0=rsin \theta[/itex]. Now, as x and y approach 0, there is no angle that [itex]\theta[/itex] approaches that make [itex]rcos \theta [/itex] and [itex]rsin \theta[/itex] simultaneously zero; hence, it must be that r approaches 0 as to make each expression zero simultaneously.
In summary, [itex]\lim_{(x,y) \rightarrow (0,0)}~(x+y) = lim_{r \rightarrow 0} ~ (rcos \theta + rsin \theta)[/itex].
Are my arguments correct?