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Question: If the normal line to the curve y = ln x at the point (u, ln u) intersects the x-axis at the point (a(u), 0), prove that $\lim \limits_{u \to \infty} (a(u) − u) = 0$.

I tried doing the question but instead of 0, I got $\infty$.

$\frac{dy}{dx}=\frac1x$ thus the gradient of normal line is -x and at u, the gradient will be -u

From the 2 points that was given, I got a gradient of $\frac{a(u)-u}{0-ln(u)}$.

As both gradient should be the same, thus I got $\frac{a(u)-u}{0-ln(u)}=-u$.

After rearranging: a(u)=u ln(u)+u

$\lim \limits_{u \to \infty} (a(u) − u) = \lim \limits_{u \to \infty}(u ln(u)+u-u) = \lim \limits_{u \to \infty}(u ln(u)) = \infty $

Did I make a mistake somewhere or is this the wrong approach?

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    $\begingroup$ The gradient of the normal is $$\frac{0 - \ln u}{a(u)-u}.$$ $\endgroup$ – Daniel Fischer Mar 26 '14 at 10:42

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