anonymous
  • anonymous
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x + h)^3 − x^3 (over) h h → 0
Mathematics
katieb
  • katieb
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SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}\) (Or the derivative of x\(^3\), for which you should get 3x\(^2\), by the power rule, thus we know what value must your limit be equal to. --> If you have ever learned the power rule)
SolomonZelman
  • SolomonZelman
You have to expand the \((x+h)^3\), at first.
anonymous
  • anonymous
\[h^3+3h^2x+3hx^2+x^3\]

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SolomonZelman
  • SolomonZelman
Yes, and now, write that on top of your fraction instead of (x+h)³.
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}\) \(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2+x^3-x^3}{h}}\)
SolomonZelman
  • SolomonZelman
\(x^3\) goes away, and then h will cancel.
SolomonZelman
  • SolomonZelman
(The validity of dividing by h on top and bottom, as h\(\rightarrow\)0, is justified by the fact that you are taking values that are not actually equal to 0, rather near 0.)
SolomonZelman
  • SolomonZelman
\(\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2\cancel{~+x^3-x^3~}}{h}}\) \(\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2}{h}}\)

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