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Math Help - Limits and Continuity

  1. #1
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    Limits and Continuity

    Prove directly: (x+1)^3 ----> 1 as x----> 0.

    So 0<mod(x-0)< delta.

    need help cheers
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  2. #2
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    Quote Originally Posted by ardam View Post
    Prove directly: (x+1)^3 ----> 1 as x----> 0.
    \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3  \leqslant \left[ {\left| x \right| + 1} \right]^3  < 8

    \varepsilon  > 0\, \Rightarrow \,\delta  = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}
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  3. #3
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    Quote Originally Posted by Plato View Post
    \left| x \right| < 1\, \Rightarrow \,\left| {\left( {x + 1} \right)^3 } \right| = \left| {\left( {x + 1} \right)} \right|^3  \leqslant \left[ {\left| x \right| + 1} \right]^3  < 8

    \varepsilon  > 0\, \Rightarrow \,\delta  = \min \left\{ {1,\frac{\varepsilon }{8}} \right\}
    So you used the condition \delta < (and equal to) 1
    To obtain [tex]\left| x \right| < 1\ which means x is in distance of one.
    So by putting x = 1 into (x+1)^3 you got 8.
    Is this correct?
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  4. #4
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    Quote Originally Posted by ardam View Post
    So you used the condition \delta < (and equal to) 1
    To obtain [tex]\left| x \right| < 1\ which means x is in distance of one.
    So by putting x = 1 into (x+1)^3 you got 8.
    Is this correct?
    Not really. I was trying to show you how to think about it.
    \left| {\left( {x + 1} \right)^3  - 1} \right| = \left| x \right|\left| {\left( {x + 1} \right)^2  + \left( {x + 1} \right) + 1} \right|

    Now use the ideas in the other post to bound the factor \left| {\left( {x + 1} \right)^2  + \left( {x + 1} \right) + 1} \right|
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