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LaddiusMaximus

  • 4 years ago

determine if this limit exists, if so evaluate it. lim h-->0 (2/(4-h)^2-2/16)/h how do you determine if it exists?

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  1. anonymous
    • 4 years ago
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    \[\lim_{h\rightarrow 0}\frac{2}{(4-h)^2}-\frac{2}{16}\]?

  2. anonymous
    • 4 years ago
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    you actually have to do the subtraction to get \[\frac{32-2(4-h)^2}{16(4-h)^2}\] and yes, there is another h in the denominator, we can put it in later

  3. anonymous
    • 4 years ago
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    now multiply out in the numerator and get \[\frac{32-2(16-8h+h^2)}{16(4-h)^2}\] \[\frac{32-32+16h-2h^2}{16(4-h)^2}\] \[\frac{8h-h^2}{8(4-h)^2}\] now lets divide by h and get \[\frac{8-h}{8(4-h)^2}\]and finally replace h by zero to get your answer

  4. anonymous
    • 4 years ago
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    you get \[\frac{8-0}{8(4-0)^2}=\frac{8}{8\times 16}=\frac{1}{16}\]

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