# Help visualising a limit proof

1. Jan 5, 2009

### transgalactic

i tried to interpret it on a sketch

http://img181.imageshack.us/img181/6498/13549423mx3.gif [Broken]

Last edited by a moderator: May 3, 2017
2. Jan 5, 2009

### transgalactic

i cant see the animation of this proove

3. Jan 5, 2009

### HallsofIvy

When you don't know where to start, look at the definitions!

"$\lim_{x\rightarrow x_0} f(x)= L$" means, by definition, that

"Given any [\itex]\epsilon> 0[/itex] there exist $\delta> 0$ such that if $|x- x_0|< \delta$ then [/itex]|f(x)- L|< \epsilon[/itex]".

Here, the function is just f(x)= x and you want to prove that the limit is $x_0$: write exactly the same thing but replace "f(x)" with "x" and "L" with "$x_0$".

"Given any [\itex]\epsilon> 0[/itex] there exist $\delta> 0$ such that if $|x- x_0|< \delta$ then [/itex]|x- x_0|< \epsilon[/itex]".

You should see immediately that what was before "$|f(x)- L|< \epsilon$" is now "$|x- x_0|< \epsilon$" the same as with "$|x- x_0|< \delta$". So make $\epsilon$ and $\delta$ the same: given any $\delta> 0$, you can always choose $\delta= \epsilon$.

4. Jan 5, 2009

### HallsofIvy

As for your picture, it would be better to draw it in two dimensions: draw the graph of y= x, a straight line. Since [/itex]\epsilon[/itex] is bounding the value of the function, draw two horizontal lines at $Y= X+ \epsilon$ and at $y= X- \epsilon$. Draw vertical lines where the graph crosses those horizontal lines to get a "box" bounding the graph. The vertical lines will give $X+\delta$ and $x-\delta$. Here, of course, because the line is y= x, that box is a square: $\delta= \epsilon$.