Question

evaluate the limit and write your answer in simplest form:
lim_{h
ightarrow0}\frac{(-1 + h)^{3}+(-1 + h)^{2}-(-1)^{3}+(-1)^{2}}{h}

Explanation:

Step1: Expand the expressions

First, expand \((-1 + h)^3\) and \((-1 + h)^2\) using the formulas \((a + b)^3=a^3+3a^2b + 3ab^2+b^3\) and \((a + b)^2=a^2+2ab + b^2\).
\((-1 + h)^3=(-1)^3+3(-1)^2h+3(-1)h^2+h^3=-1 + 3h-3h^2+h^3\)
\((-1 + h)^2=(-1)^2+2(-1)h+h^2=1 - 2h+h^2\)
\((-1)^3=-1\), \((-1)^2 = 1\)
The numerator becomes \((-1 + 3h-3h^2+h^3+1 - 2h+h^2)-(-1 + 1)=h - 2h^2+h^3\)

Step2: Simplify the limit

The original limit \(\lim_{h
ightarrow0}\frac{[(-1 + h)^3+(-1 + h)^2]-[(-1)^3+(-1)^2]}{h}=\lim_{h
ightarrow0}\frac{h - 2h^2+h^3}{h}\)
Factor out \(h\) from the numerator: \(\lim_{h
ightarrow0}\frac{h(1 - 2h+h^2)}{h}\)
Cancel out the common - factor \(h\) (since \(h
eq0\) when taking the limit as \(h
ightarrow0\)), we get \(\lim_{h
ightarrow0}(1 - 2h+h^2)\)

Step3: Evaluate the limit

Substitute \(h = 0\) into \(1 - 2h+h^2\), we have \(1-2\times0 + 0^2=1\)

Answer:

\(1\)