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{e^x - 1 - x}{x^2} $
Solution:
$ \begin{eqnarray*} & & \lim\limits_{x \to 0} \frac{e^x - 1 - x}{x^2} \\ &=& \lim\limits_{x \to 0} \frac{e^x - 1}{2x} \\ &=& \lim\limits_{x \to 0} \frac{e^x}{2} \\ &=& \frac{1}{2} \\ \end{eqnarray*} $
Again, this is easy to validate 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{e^x - 1 - x}{x^2} $
Solution:
$ \begin{eqnarray*} & & \lim\limits_{x \to 0} \frac{e^x - 1 - x}{x^2} \\ &=& \lim\limits_{x \to 0} \frac{e^x - 1}{2x} \\ &=& \lim\limits_{x \to 0} \frac{e^x}{2} \\ &=& \frac{1}{2} \\ \end{eqnarray*} $
Again, this is easy to validate numerically:
|
x
|
f(x)
|
$ e^2 $
|
|
0.1
|
0.517091808
|
0.000292
|
|
0.01
|
0.501670842
|
2.79E-06
|
|
0.001
|
0.500166708
|
2.78E-08
|
|
0.0001
|
0.500016671
|
2.78E-10
|
|
0.00001
|
0.500000696
|
4.85E-13
|
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