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anonymous

  • one year ago

Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x + h)^3 − x^3 (over) h h → 0

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  1. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}\) (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)

  2. SolomonZelman
    • one year ago
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    You have to expand the \((x+h)^3\), at first.

  3. anonymous
    • one year ago
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    \[h^3+3h^2x+3hx^2+x^3\]

  4. SolomonZelman
    • one year ago
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    Yes, and now, write that on top of your fraction instead of (x+h)³.

  5. SolomonZelman
    • one year ago
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    \(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}\) \(\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2+x^3-x^3}{h}}\)

  6. SolomonZelman
    • one year ago
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    \(x^3\) goes away, and then h will cancel.

  7. SolomonZelman
    • one year ago
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    (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.)

  8. SolomonZelman
    • one year ago
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    \(\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2\cancel{~+x^3-x^3~}}{h}}\) \(\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2}{h}}\)

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