Limits: finding limits with triangles in them

[volleyball_freak31]

How do you find a limit with the triangles in it?

for example:
lim as triangle x approaches 0.

[(x + triangle x)^2 - 2(x + triangle x) + 1 - (x^2 - 2x +1)] / triangle x

 

[galactus]

Try simplifying first.

\(\displaystyle \L\\(x+{\Delta}x)^{2}-2(x+{\Delta}x)+1-(x^{2}-2x+1)=2x{\Delta}x+({\Delta}x)^{2}-2({\Delta}x)\)

\(\displaystyle \L\\\frac{2x{\Delta}x+({\Delta}x)^{2}-2{\Delta}x}{{\Delta}x}\)

=\(\displaystyle \L\\2x+{\Delta}x-2\)

\(\displaystyle \L\\\lim_{{\Delta}x\to\0}2x+{\Delta}x-2\)

 

[soroban]

Re: Limits

Hello, volleyball_freak31!

How do you find a limit with the triangles in it?

For example: \(\displaystyle \L\,\lim_{\Delta x\to0}\frac{(x\,+\,\Delta x)^2 - 2(x\,+\,\Delta x)\,+\,1 \,-\,(x^2\,-\,3x\,+\,1)}{\Delta x}\)

Obviously, you are working with the definition of a derivative.


That \(\displaystyle \Delta x\) ("delta x") is a single term . . . replace it with \(\displaystyle h\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \L\,\lim_{h\to0}\frac{(x\,+\,h)^2\,-\,2(x\,+\,h)\,+\,1\,-\,(x^2\,-\,2x\,+\,1)}{h}\)


Does that look familiar?