Question

Find the limit: 20%2B%20ax%29%20%2B%201%20-%20%28x%5E%7B2%7D%20-%202x%20%2B%201%29%7D%7Bax%7D" id="TexFormula1" title="\lim_{a x \to 0} \frac{(x + ax)^{2}-2(x + ax) + 1 - (x^{2} - 2x + 1)}{ax}" alt="\lim_{a x \to 0} \frac{(x + ax)^{2}-2(x + ax) + 1 - (x^{2} - 2x + 1)}{ax}" align="absmiddle" class="latex-formula">

Answer 1

Answer:

2x-2

Step-by-step explanation:

lim ax goes to 0   ( x+ ax)^2 -2 ( x+ax) +1 - ( x^2 -2x+1)

                              --------------------------------------------------

                                                    ax

Simplify the numerator  by foiling the first term and distributing the minus signs

 x^2+ 2ax^2 + a^2 x^2 -2x-2ax +1 -  x^2 +2x-1

 --------------------------------------------------

                   ax

Combine like terms

2ax^2 + a^2 x^2 -2ax  

 --------------------------------------------------

                   ax

Factor out ax

ax( 2x + ax -2)

----------------------

     ax

Cancel ax

2x + ax -2

Now take the limit

lim ax goes to 0  ( 2x + ax -2)

                                  2x +0-2

                                   2x -2

Answer 2

I'll let h = ax, so the limit is

$\displaystyle\lim_{h\to0}\frac{(x+h)^2-2(x+h)+1-(x^2-2x+1)}h$

i.e. the derivative of $x^2-2x+1$.

Expand the numerator to see several terms that get eliminated:

$(x+h)^2-2(x+h)+1-(x^2-2x+1)=x^2+2xh+h^2-2x-2h+1-x^2+2x-1=2xh+h^2-2h$

So we have

$\displaystyle\lim_{h\to0}\frac{2xh+h^2-2h}h$

Since h ≠ 0 (because it is approaching 0 but never actually reaching 0), we can cancel the factor of h in both numerator and denominator, then plug in h = 0:

$\displaystyle\lim_{h\to0}(2x+h-2)=\boxed{2x-2}$