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hooverst

  • 2 years ago

Calculate the following limit: lim h→0 (4 + h)3− 64 over h

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  1. Eyad
    • 2 years ago
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    Ok Here is your problem : \[\Large lim h-->0 \frac{(f+h)^3-64}{h}\]

  2. Eyad
    • 2 years ago
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    64=4^3

  3. campbell_st
    • 2 years ago
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    expand the cubic \[\lim_{h \rightarrow 0}\frac{64 + 48h + 12h^2 + h^3 - 64}{h}\] which simplifes to \[\lim_{h \rightarrow 0}\frac{48h + 12h^2 + h^3}{h}\] take out the common factor of h leaves \[\lim_{h \rightarrow 0} 48 + 12h + h^2\] give the limit of 48

  4. Eyad
    • 2 years ago
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    Here is a Hint : ______________ Add 4 and subtract 4 from the denominator : it will be \[\Large \frac{(4+h)^3-4^3}{(4+h)-4}\] Now its easy : The answer is 3/1*4^(3-1)=48

  5. hooverst
    • 2 years ago
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    thanks yall

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