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hw5 the limit laws (target l4; §2.3)
score: 4/13 answered: 4/13
question 5
evaluate the limit: $lim_{h
ightarrow0}\frac{(9 + h)^{2}-81}{h}$
question help: video message instructor

Explanation:

Step1: Expand the numerator

Expand \((9 + h)^2-81\). Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), where \(a = 9\) and \(b=h\), we get \((9 + h)^2-81=(81 + 18h+h^{2})-81=18h + h^{2}\).

Step2: Simplify the fraction

The original limit \(\lim_{h
ightarrow0}\frac{(9 + h)^2-81}{h}\) becomes \(\lim_{h
ightarrow0}\frac{18h+h^{2}}{h}\). Factor out an \(h\) from the numerator: \(\lim_{h
ightarrow0}\frac{h(18 + h)}{h}\). Cancel out the common - factor \(h\) (since \(h
eq0\) when taking the limit), and we have \(\lim_{h
ightarrow0}(18 + h)\).

Step3: Evaluate the limit

Substitute \(h = 0\) into \(18 + h\). We get \(18+0 = 18\).

Answer:

18