Tangent of function and its limit position

[gruba]

Homework Statement

Find tangent line of y=xe^{\frac{1}{x}} at point x=\alpha and it's limit position when \alpha \rightarrow +\infty.

Homework Equations

Tangent of y=f(x) at point M(x_0,f(x_0)): y-y_0=f^{'}(x_0)(x-x_0)

The Attempt at a Solution

Applying the above equation for tangent of function,
y_0=\alpha e^{\frac{1}{\alpha}}, f^{'}(x_0)=\frac{(\alpha-1)e^{\frac{1}{\alpha}}}{\alpha}, x_0=\alpha
gives
y-\alpha e^{\frac{1}{\alpha}}=\frac{(\alpha-1)e^{\frac{1}{\alpha}}}{\alpha}(x-\alpha)

How to find limit position of a tangent? Is it a limit of y when \alpha \rightarrow +\infty?

 

[andrewkirk]

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.

 

[RUber]

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.