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Study23
Please help me with this limit!!!
\(\ \Huge \lim_{h \rightarrow 0} \ \frac{(2+h)^3-8}{h} .\) Do I just multiply h/h, to get 1 on the denominator??
then subtract 8 and the usual miracle will occur everything without an \(h\) will be gone, so you can factor an \(h\) out of the numerator, cancel with the \(h\) in the denominator, then replace \(h\) by 0
How do I expand the trinomial?
if you do not know that \((a+b)^2=a^3+3a^2b+3ab^2+a^3\) then you have the annoying task of multiplying by hand \[(2+h)(2+h)(2+h)\]
one more time \[(a+b)^3=a^3+3a^2b+3ab^2+b^3\]