Question

use algebra to evaluate the limit $lim_{h \to 0}\frac{(5 - h)^{3}-125}{h}=$

Explanation:

Step1: Expand $(5 - h)^3$

Using the formula $(a - b)^3=a^3-3a^2b + 3ab^2 - b^3$, where $a = 5$ and $b=h$, we have $(5 - h)^3=5^3-3\times5^2\times h+3\times5\times h^2 - h^3=125 - 75h+15h^2 - h^3$.

Step2: Substitute the expansion into the limit

The original limit $\lim_{h
ightarrow0}\frac{(5 - h)^3-125}{h}$ becomes $\lim_{h
ightarrow0}\frac{125 - 75h+15h^2 - h^3-125}{h}$.

Step3: Simplify the numerator

The numerator $125 - 75h+15h^2 - h^3-125=-75h + 15h^2 - h^3$. So the limit is $\lim_{h
ightarrow0}\frac{-75h + 15h^2 - h^3}{h}$.

Step4: Factor out $h$ from the numerator

$\frac{-75h + 15h^2 - h^3}{h}=h\frac{-75 + 15h - h^2}{h}=-75 + 15h - h^2$ for $h
eq0$.

Step5: Evaluate the limit

Now, $\lim_{h
ightarrow0}(-75 + 15h - h^2)=-75+15\times0 - 0^2=-75$.

Answer:

$-75$