Scientific Computing - Quiz 1 - Question 2

Problem:

L'Hopital's rule is a method for determining the value of indeterminate forms. Determine the value of the following limits:

$\lim\limits_{x \to 0} \frac{2\sin x - \sin 2x}{x-\sin x}$

Solution:

$\begin{eqnarray*} & & \lim\limits_{x \to 0} \frac{2\sin x - \sin 2x}{x - \sin x} \\ &=& \lim\limits_{x \to 0} \frac{2\cos x - 2 \cos 2x}{1 -\cos x} \\ &=& \lim\limits_{x \to 0} \frac{-2\sin x + 4 \sin 2x}{\sin x} \\ &=& \lim\limits_{x \to 0} \frac{-2\sin x + 8 \sin x \cos x}{\sin x} \\ &=& \lim\limits_{x \to 0} -2 + 8 \cos x \\ &=& 6 \end{eqnarray*}$

One can easily verify this numerically:

x	f(x)	$e^2$
0.1	5.988008283	0.000144
0.01	5.999880001	1.44E-08
0.001	5.9999988	1.44E-12
0.0001	6.000000163	2.64E-14
0.00001	5.999979671	4.13E-10