Limit $\frac{f(x)}{g(x)}$: Solve w/ L'H Rule

In summary: This is just one example, but there are infinitely many other choices for the constants that would also work. Essentially, you can choose any constant that makes the functions vanish when $x \to 0$. So in summary, in order for the L'H rule to be correctly applied twice, the constants of the functions $f(x)$ and $g(x)$ must be chosen so that the functions vanish when $x\to0$, which allows for an infinite number of possible choices.
  • #1
karush
Gold Member
MHB
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Consider the following limit where L'H Rule was correctly applied twice
Determine the functions f'(x), g'(x), f(x), and g(x) needed to result in the limit given.
\begin{align*}\displaystyle
\lim_{x \to 0}\frac{f(x)}{g(x)}
\overset{\text{L'H}}=&
\lim_{x \to 0}\frac{f'(x)}{g'(x)}\\
\overset{\text{L'H}}=&
\lim_{x \to 0}\frac{f''(x)}{g''(x)}\\
\overset{\text{}}=&
\lim_{x \to 0}\frac{e^x + \cos{x}}{12x-6}\\
\overset{\text{}}=&
\lim_{x \to 0}\frac{e^0 + \cos{0}}{12(0)-6}
=\frac{1+1}{-6}=\frac{1}{3}\\
\end{align*}
$ \textit{using}$
$$\displaystyle \int \cos{x} \, dx = \sin{x}+c
,\quad
\displaystyle \int \sin{x} \, dx=-\cos{x}+c
,\quad
\displaystyle \int e^x \, dx = e^x+c
,\quad
\displaystyle \int x^n \, dx =\frac{x^{n+1}}{n+1}+c$$
$\textit{then}$
\begin{align*}\displaystyle
\frac{f'(x)}{g'(x)}
&=\frac{e^x-\sin{x}}{6x^2-6x}\\
\frac{f(x)}{g(x)}
&=\frac{e^x + \cos{x}}{2x^3-3x^2}
\end{align*}ok not sure if there are typos in this
but was not sure how deal with the constant c with anti-direvatives
 
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  • #2
There is one typo: [tex]\frac{1+ 1}{-6}= -\frac{1}{3}[/tex], not [tex]\frac{1}{3}[/tex].

As for the constants, there are an infinite number of functions, f and g, that will work here.

With [tex]f''(x)= e^x+ sin(x)[/tex], [tex]f'(x)= e^x- cos(x)+ A[/tex] and, integrating again, [tex]f(x)= e^x- sin(x)+ Ax+ B[/tex] where A and B are arbitrary constants. With [tex]g''(x)= 12x- 6[/tex], [tex]g'(x)= 6x^2- 6x+ C[/tex] and, [tex]g(x)= 2x^3- 2x^2+ Cx+ D[/tex] where C and D are arbitrary constants.
 
  • #3
Country Boy said:
As for the constants, there are an infinite number of functions, f and g, that will work here.
But in order for the L'H rule to apply, the constants must be chosen so that the functions vanish when $x\to0$.
 
  • #4
how about just using 0 for placeholder constant
 
Last edited:
  • #5
karush said:
how about just using 0 for placeholder constant
In your example, if $f''(x) = e^x + \cos x$ then $f'(x) = e^x + \sin x + \text{const.}$ To ensure that $f'(x) = 0$ you need to take the constant to be $-1$. So $f'(x) = e^x + \sin x -1$ and then $f(x) = e^x - \cos x - x + \text{const.}$ That vanishes when $x=0$ if the constant is $0$, so the conclusion is that $f(x) = e^x - \cos x - x$.
 

FAQ: Limit $\frac{f(x)}{g(x)}$: Solve w/ L'H Rule

What is the Limit $\frac{f(x)}{g(x)}$?

The limit of $\frac{f(x)}{g(x)}$ is the value that the function approaches as the input (x) approaches a certain point. It is represented using the notation $\lim_{x \to a} \frac{f(x)}{g(x)}$.

What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical theorem that allows us to evaluate limits involving indeterminate forms, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. It states that if the limit of $\frac{f(x)}{g(x)}$ is an indeterminate form, then the limit can be evaluated by taking the derivative of both the numerator and denominator and then re-evaluating the limit.

When should L'Hopital's Rule be used?

L'Hopital's Rule should be used when evaluating a limit of the form $\frac{f(x)}{g(x)}$, where both f(x) and g(x) approach 0 or infinity as x approaches a certain point. It can also be used when the limit results in an indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

How do you apply L'Hopital's Rule?

To apply L'Hopital's Rule, take the derivative of both the numerator and denominator of the original limit. Then, re-evaluate the limit using these new derivatives. If the resulting limit is still an indeterminate form, repeat the process until a non-indeterminate form is obtained or until it is clear that the limit does not exist.

Are there any restrictions when using L'Hopital's Rule?

Yes, there are some restrictions when using L'Hopital's Rule. First, the limit must be in the form of $\frac{f(x)}{g(x)}$. Second, both the numerator and denominator must approach 0 or infinity as x approaches the given point. Third, the limit must result in an indeterminate form, such as $\frac{0}{0}$ or $\frac{\infty}{\infty}$. If these conditions are not met, then L'Hopital's Rule cannot be applied.

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