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Math Help - Specific Point of Limit on graph

  1. #1
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    Specific Point of Limit on graph

    Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.

    lim[(1+h)^(1/6)-1]/h
    h->0

    I evaluated the limit to be
    M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.

    The answer in the back says:
    M_t=1/6 to y=x^(1/6) at x=1

    how did they get x=1?
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.
    lim[(1+h)^(1/6)-1]/h
    h->0
    I evaluated the limit to be
    M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.
    The answer in the back says:
    M_t=1/6 to y=x^(1/6) at x=1
    how did they get x=1?
    Actually you are given that \color{red}x=1.
    It is part of the statement of the problem. See above.
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  3. #3
    Senior Member DeMath's Avatar
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    Quote Originally Posted by skeske1234 View Post
    Interpret the limit as the slope of the tangent line to a curve at a specific point after you evaluate the limit.

    lim[(1+h)^(1/6)-1]/h
    h->0

    I evaluated the limit to be
    M_t=1/6 to y=x^(1/6) but I am having trouble finding the x co-ordinate.

    The answer in the back says:
    M_t=1/6 to y=x^(1/6) at x=1

    how did they get x=1?
    If you need evaluate the \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/6}} - 1}}{h} then, using the formula for the difference of cubes, we get

    {\left( {1 + h} \right)^{1/2}} - 1 = \left( {{{\left( {1 + h} \right)}^{1/6}} - 1} \right)\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right) \Leftrightarrow

    \Leftrightarrow {\left( {1 + h} \right)^{1/6}} - 1 = \frac{{{{\left( {1 + h} \right)}^{1/2}} - 1}}{{{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1}}.

    So we have

    \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/6}} - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{1/2}} - 1}}{{h\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)}} =

    = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {1 + h} \right)}^{1/2}} - 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}}{{h\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}} =

    = \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {{{\left( {1 + h} \right)}^{1/3}} + {{\left( {1 + h} \right)}^{1/6}} + 1} \right)\left( {{{\left( {1 + h} \right)}^{1/2}} + 1} \right)}} = \frac{1}{{3 \cdot 2}} = \frac{1}{6}.
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  4. #4
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    Put t=\sqrt[6]{h+1} so that t\to1 and h=t^6-1.
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