# L'Hopital / Second Derivative

1. Sep 9, 2004

### IrishGuy

The teacher wrote the solution to the problem on the board without giving much explanation of how he got there, and now he wants the third derivative, so I was wondering if you could help answer how he got the second.

g(x) [the lim. as x goes to zero] = (e^x - 1) / x , where x can not equal zero

then he wrote

1, x = 0 I'm guessing this means to use the value 1 when x = 0

Then he wrote the first derivative of g(x) is

( xe^x - e^x + 1) / (x^2) , where x does not equal zero

and

1/2 x = 0

then the second derivative of g(x) :

g(x) [lim. as x goes to zero] ( ( (xe^x -e^x + 1) / x^2 ) - (1/2) ) / x

then goes to:

( 2xe^x - 2e^x + 2 - x^2 ) / 2x^3

then goes to:

( 2e^x + 2xe^x - 2e^x - 2x) / 6x^2 [the 2e^x's cancel]

this goes to:

( e^x - 1 ) / 3x = 1/3

Now I have to do the third derivative. If you could help me understand this second derivative it may help me solve the third. Thanks in advance for any time you spend on this.

-Kevin

2. Sep 11, 2004

### arildno

I can't say I like your teacher's method overly much, although it is correct.
Let's first write g(x) in a form where it is easy to find the values of g and g's derivatives at zero:
1. The exponential series.
Possibly, you haven't learnt this yet, but for any x we have:
$$e^{x}=1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}++$$
where the INFINITE SERIES can be written as:
$$e^{x}=\sum_{i=0}^{\infty}\frac{x^{i}}{i!}$$
where 0!=1, and i!=1*2..*i (i>0)
(The factorial)
2. Rewriting g(x)
From 1., and the expression for g(x), we get:
$$g(x)=\frac{\sum_{i=0}^{\infty}\frac{x^{i}}{i!}-1}{x}=\sum_{i=1}^{\infty}\frac{x^{i-1}}{i!}$$
3.
The first terms of g(x) can be written out as:
g(x)=1+\frac{1}{2}x+\frac{1}{6}x^{2}+\frac{1}{24}x^{3}+++[/tex]

In order to find g(0),g'(0),g''(0),g'''(0), it is sufficient to use the 3.degree polynomial appoximation to g(x):
$$g_{p}(x)=1+\frac{1}{2}x+\frac{1}{6}x^{2}+\frac{1}{24}x^{3}$$
You gain:
$$g(0)=g_{p}(0)=1$$
$$g'(0)=g_{p}'(0)=\frac{1}{2}$$
$$g''(0)=g_{p}''(0)=\frac{1}{3}$$
$$g'''(0)=g_{p}'''(0)=\frac{1}{4}$$