Question

Answer it:::
$\lim_{\Delta x \to 0} \dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}$​

Answer 1

Answer:

$\red{\lim_{\Delta x \to 0} \dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}\implies{2x-2}}$

Step-by-step explanation:

$\mapsto$So we want to find the limit:

$\lim_{\Delta x \to 0} \dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}$

$\hookrightarrow$Expand the numerator:,

$\lim_{\Delta x \to 0} \dfrac{(x^2+2x\Delta x+\Delta x^2)+(-2x-2\Delta x)+1+(-x^2+2x-1)}{\Delta x}$

$\mapsto$Simplify the numerator. Combine like terms:,

$\lim_{\Delta x\to 0}\dfrac{(x^2-x^2)+(-2x+2x)+(1-1)+(2x\Delta x+\Delta x^2-2\Delta x)}{y}$

$\mapsto$The first term terms cancel:

$\lim_{\Delta x\to 0} \dfrac{2x\Delta x+\Delta x^2-2\Delta x}{\Delta x}$

$\mapsto$Factor out a Δx:

$\lim_{\Delta x \to 0} \dfrac{\Delta x(2x+\Delta x-2)}{\Delta x}$

$\mapsto$The Δxs cancel:

$\lim_{\Delta x\to 0}2x+\Delta x-2$

$\mapsto$Now, use direct substitution:

$\begin{gathered}2x+(0)-2\\=2x-2\end{gathered}$

Therefore :

$\blue{\lim_{\Delta x \to 0} \dfrac{(x+\Delta x)^2-2(x+\Delta x)+1-(x^2-2x+1)}{\Delta x}=2x-2}$

Answer 2

EXPLANATION.

$\sf \implies \lim_{\Delta x \to 0} \dfrac{(x + \Delta x)^{2} - 2(x + \Delta x)+ 1 - (x^{2}- 2x + 1) }{\Delta x}$

As we know that,

First we can put the value of Δx in equation,

And check in which form the equation is.

$\sf \implies \lim_{\Delta x \to 0} \dfrac{(x + 0)^{2} - 2(x + 0)+ 1 - (x^{2}- 2x + 1) }{0}.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{(x)^{2} - 2x+ 1 - x^{2}+ 2x - 1) }{0}.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{0}{0}.$

As we can see it is in the form of 0/0,

We can simply equate (factorizes) this equation, we get.

$\sf \implies \lim_{\Delta x \to 0} \dfrac{(x^{2} + \Delta x^{2} + 2x\Delta x ) + (- 2x - 2 \Delta x) + 1 - (x^{2} - 2x + 1)}{\Delta x}.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{x^{2} + 2x \Delta x + \Delta x^{2} - 2x - 2 \Delta x + 1 - x^{2} + 2x - 1}{\Delta x}.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{2x \Delta x + \Delta x^{2} - 2 \Delta x}{\Delta x}.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{\Delta x(2x + \Delta x - 2)}{\Delta x}.$

Put the value of Δx = 0 in equation, we get.

$\sf \implies \lim_{\Delta x \to 0} 2x + 0 - 2.$

$\sf \implies \lim_{\Delta x \to 0} 2x - 2.$

$\sf \implies \lim_{\Delta x \to 0} \dfrac{(x + \Delta x)^{2} - 2(x + \Delta x)+ 1 - (x^{2}- 2x + 1) }{\Delta x} = 2x - 2.$

Answer = (2x - 2).