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:
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)
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$e^2$
|
|
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
|
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