Question

lim deltax>0 (x+deltax)^2-2(x+deltax)+1- (x^2-2x+1)/deltax

Answer 1

\(\bf lim_{\Delta \to 0}\quad \cfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}\quad ?\)

Answer 2

\[\bf lim_{\Delta x\to 0}\quad \cfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}\quad \]

Answer 3

lots of algebra to do in the numerator
i would replace \(\Delta x\) by \(h\) to make life easier and write
\[\lim_{h\to 0}\frac{(x+h)^2-2(x+h)+1-(x^2-2x+1)}{h}\]

Answer 4

the numerator is
\[x^2+2xh+h^2-2x-2h+1-x^2+2x-1\] and as usual everything without an \(h\) in it adds up to zero leaving onlly
\[2xh-2h+h^2\]

Answer 5

divide by \(h\) and get
\[2x-2+h^2\]

Answer 6

but the thing is that a need to simplifly to the point that it can´t anymore and then replace the h for 0 , so a just replace the h of \[2x-2+h ^{2}\] and the problem will be over?

Answer 7

\(\bf lim_{\Delta \to 0}\quad \cfrac{({\color{brown}{ x+\Delta x}})^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}
\\ \quad \\
\implies \cfrac{({\color{brown}{ x^2+2\Delta x+\Delta^2}})-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}
\\ \quad \\
\implies \cfrac{\cancel{ x^2 }\cancel{ +2\Delta x }+{\color{brown}{ \cancel{ \Delta }}}^2
\cancel{ -2x }\cancel{ -2\Delta x }\cancel{ +1 }\cancel{ -x^2 }\cancel{ +2x }\cancel{ -1 })}
{{\color{brown}{ \cancel{ \Delta }}} x}\)