How Does L'Hopital's Rule Apply to the Second Derivative Calculation?

[IrishGuy]

Any help you can give me would be appreciated!

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

 

[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 learned this yet, but for any x we have:
[tex]e^{x}=1+x+\frac{1}{2}x^{2}+\frac{1}{6}x^{3}+\frac{1}{24}x^{4}++[/tex]
where the INFINITE SERIES can be written as:
[tex]e^{x}=\sum_{i=0}^{\infty}\frac{x^{i}}{i!}[/tex]
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:
[tex]g(x)=\frac{\sum_{i=0}^{\infty}\frac{x^{i}}{i!}-1}{x}=\sum_{i=1}^{\infty}\frac{x^{i-1}}{i!}[/tex]
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):
[tex]g_{p}(x)=1+\frac{1}{2}x+\frac{1}{6}x^{2}+\frac{1}{24}x^{3}[/tex]
You gain:
[tex]g(0)=g_{p}(0)=1[/tex]
[tex]g'(0)=g_{p}'(0)=\frac{1}{2}[/tex]
[tex]g''(0)=g_{p}''(0)=\frac{1}{3}[/tex]
[tex]g'''(0)=g_{p}'''(0)=\frac{1}{4}[/tex]

 

[M98Ranger]

Sure, I'd be happy to help you understand the second derivative of g(x) and how it relates to the third derivative. First, let's review the concept of L'Hopital's rule. This rule states that if you have a limit of the form 0/0 or ∞/∞, you can take the derivative of the numerator and denominator separately and evaluate the limit again. This process can be repeated as many times as necessary until a non-indeterminate form is reached. This is a useful tool for evaluating limits that involve fractions.

Now, let's look at the second derivative of g(x). You correctly wrote out the first derivative as (xe^x - e^x + 1)/(x^2), where x≠0. When finding the second derivative, we apply L'Hopital's rule again because we have a limit of the form 0/0. This means we take the derivative of the numerator and denominator separately and evaluate the limit again. However, in this case, we also have a constant term, 1/2, in the numerator. This means we need to take the derivative of that as well. So, the second derivative becomes:

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

= (2xe^x + e^x - 1 - 1/2x^2) / x^3

= (2e^x + e^x - 1 - x) / 2x^2

= (3e^x - 1 - x) / 2x^2

Now, to find the third derivative, we can apply L'Hopital's rule again because we have a limit of the form 0/0. This means we take the derivative of the numerator and denominator separately and evaluate the limit again. The third derivative becomes:

g(x) [lim. as x goes to zero] ((3e^x - 1 - x)/2x^2 - 1/3) / x

= (3e^x - 1 - x - 2/3x^3) / 2x^3

= (3e^x - 1 - x - 2/3x^3) / 2x^3

= (3e^x