Answer:
[tex]\displaystyle \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}=2x-2[/tex]
Step-by-step explanation:
We want to find the limit:
[tex]\displaystyle \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}[/tex]
We can expand the numerator:
[tex]=\displaystyle \lim_{\Delta x \to 0} \frac{(x^2+2x\Delta x+\Delta x^2)+(-2x-2\Delta x)+1+(-x^2+2x-1)}{\Delta x}[/tex]
Simplify. Combine like terms:
[tex]\displaystyle \lim_{\Delta x\to 0} \frac{(x^2-x^2)+(-2x+2x)+(1-1)+(2x\Delta x+\Delta x^2-2\Delta x)}{y}[/tex]
The first three terms will cancel:
[tex]\displaystyle \lim_{\Delta x\to 0} \frac{2x\Delta x+\Delta x^2-2\Delta x}{\Delta x}[/tex]
Factor:
[tex]\displaystyle \lim_{\Delta x \to 0} \frac{\Delta x(2x+\Delta x-2)}{\Delta x}[/tex]
Cancel:
[tex]\displaystyle \lim_{\Delta x\to 0}2x+\Delta x-2[/tex]
Now, we can use direct substitution:
[tex]\displaystyle \begin{aligned} &\Rightarrow 2x+(0)-2\\ &=2x-2\end{aligned}[/tex]
Therefore:
[tex]\displaystyle \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}=2x-2[/tex]
