for which of the following values of c would we get

Lim (Ln(cos cx))/ 2x^2 = -1 ?

x->0

<a>1

<b> 2

<c> 4

<d> 8

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- May 25th 2006, 12:21 PM #1

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- Sep 2005
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- May 25th 2006, 01:28 PM #2
There are several ways you could tackle this.

L'Hopitals rule equates the limit of the ratio f/g with that of the derivatives f'/g' under suitable conditions on f and g. Here the derivative of log(cos(cx)) is -c.sin(cx)/cos(cx) and the derivative of 2x^2 is 4x. Using the well known limit of sin(x)/x, written as sinc(cx)/(cx) we have -c/(4/c) = -c^2/4. This is -1 for c=2.

A numerical check: for x = 0.001, the ratio is -1.000001.