Question

Find the limit. (If an answer does not exist, enter DNE.) lim Δx→0 (x + Δx)2 − 3(x + Δx) + 2 − (x2 − 3x + 2) Δx

Answer 1

$\displaystyle\lim_{\Delta x\to0}\frac{(x+\Delta x)^2-3(x+\Delta x)+2-(x^2-3x+2)}{\Delta x}$

Expand the numerator as

$x^2+2x\Delta x+(\Delta x)^2-3x-3\Delta x+2-x^2+3x-2=(2x-3)\Delta x+(\Delta x)^2$

Then in the limit,

$\displaystyle\lim_{\Delta x\to0}\frac{(2x-3)\Delta x+(\Delta x)^2}{\Delta x}=\lim_{\Delta x\to0}(2x-3+\Delta x)=\boxed{2x-3}$

Answer 2

The value of the limit is: 2x - 3.

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• To make the typing easier, I am going to use $\Delta_x$ as h.
• Thus, we want to find the following limit:

$\lim_{h \rightarrow 0} \frac{(x + h)^2 - 3(x + h) + 2 - (x^2 -3x + 2)}{h}$

Expanding the operations:

$\lim_{h \rightarrow 0} \frac{x^2 + 2xh + h^2 - 3x - 3h + 2 - x^2 + 3x - 2}{h}$

$\lim_{h \rightarrow 0} \frac{2xh - 3h + h^2}{h}$

$\lim_{h \rightarrow 0} \frac{h(2x - 3 + h)}{h}$

Simplifying by h:

$\lim_{h \rightarrow 0} 2x - 3 + h$

Replacing h by 0:

$\lim_{h \rightarrow 0} 2x - 3 + h = 2x - 3$

The value of the limit is 2x - 3.

A similar problem is given at https://brainly.com/question/23882529