# How do you find the limit lim_(h->0)((2+h)^3-8)/h ?

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2
Mar 7, 2018

$12$

#### Explanation:

We know that,$\textcolor{red}{{\lim}_{x \to a} \frac{{x}^{n} - {a}^{n}}{x - a} = n \cdot {a}^{n - 1}}$
$L = {\lim}_{h \to 0} \frac{{\left(2 + h\right)}^{3} - 8}{h}$,let,$2 + h = x \Rightarrow h \to 0 , t h e n , x \to 2$
So,$L = {\lim}_{x \to 2} \left(\frac{{x}^{3} - {2}^{3}}{x - 2}\right) = 3 {\left(2\right)}^{3 - 1} = 3 \cdot {2}^{2} = 12$

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1
Mar 7, 2018

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Then teach the underlying concepts
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1
Jake M. Share
Mar 7, 2018

12

#### Explanation:

We can expand the cube:
${\left(2 + h\right)}^{3} = 8 + 12 h + 6 {h}^{2} + {h}^{3}$

Plugging this in,

${\lim}_{h \rightarrow 0} \frac{8 + 12 h + 6 {h}^{2} + {h}^{3} - 8}{h} = {\lim}_{h \rightarrow 0} \frac{12 h + 6 {h}^{2} + {h}^{3}}{h}$

$= {\lim}_{h \rightarrow 0} \left(12 + 6 h + {h}^{2}\right) = 12$.

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