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Homework Help: Tangent of function and its limit position

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data
    Find tangent line of [itex]y=xe^{\frac{1}{x}}[/itex] at point [itex]x=\alpha[/itex] and it's limit position when [itex]\alpha \rightarrow +\infty[/itex].

    2. Relevant equations
    Tangent of [itex]y=f(x)[/itex] at point [itex]M(x_0,f(x_0))[/itex]: [tex]y-y_0=f^{'}(x_0)(x-x_0)[/tex]

    3. The attempt at a solution
    Applying the above equation for tangent of function,
    [tex]y_0=\alpha e^{\frac{1}{\alpha}}, f^{'}(x_0)=\frac{(\alpha-1)e^{\frac{1}{\alpha}}}{\alpha}, x_0=\alpha[/tex]
    [tex]y-\alpha e^{\frac{1}{\alpha}}=\frac{(\alpha-1)e^{\frac{1}{\alpha}}}{\alpha}(x-\alpha)[/tex]

    How to find limit position of a tangent? Is it a limit of [itex]y[/itex] when [itex]\alpha \rightarrow +\infty[/itex]?
  2. jcsd
  3. Sep 21, 2015 #2


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    The question is not expressed very clearly. I am not aware of any concept of a 'limit position' of a line.

    However, from the context, one can make a pretty confident guess that the question is trying to ask if there is a line to which the tangent lines become asymptotically 'closer' (in some as yet undefined sense) as ##\alpha\to\infty##.

    A line is fully defined by its x intercept and its gradient. Does the ##\lim_{\alpha\to\infty}## of the y intercept and/or the gradient exist? If so, what are they? If you use them to define a line, that may be the line that the teacher is looking for.
  4. Sep 21, 2015 #3


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    Homework Helper

    I agree with Andrewkirk. You should rearrange the equation from point-slope form into slope-intercept form. Then you will be able to clearly see what happens to the slope and y-intercept as alpha goes to infinity.

    For your slope, ##\lim_{\alpha \to \infty} \frac{(\alpha -1)}{\alpha} e^{\frac1\alpha}## should be pretty clear if you look at the fraction part separate from the exponential part.
    For your y-intercept, once you rearrange, that should become pretty clear as well.
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