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Svetach [21]
3 years ago
14

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

Mathematics
1 answer:
yawa3891 [41]3 years ago
3 0

\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}

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