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

[karush]

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

 

[HOI]

There is one typo: $$\frac{1+ 1}{-6}= -\frac{1}{3}$$, not $$\frac{1}{3}$$.

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

With $$f''(x)= e^x+ sin(x)$$, $$f'(x)= e^x- cos(x)+ A$$ and, integrating again, $$f(x)= e^x- sin(x)+ Ax+ B$$ where A and B are arbitrary constants. With $$g''(x)= 12x- 6$$, $$g'(x)= 6x^2- 6x+ C$$ and, $$g(x)= 2x^3- 2x^2+ Cx+ D$$ where C and D are arbitrary constants.

 

[Opalg]

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 $x\to0$.

 

[karush]

how about just using 0 for placeholder constant

 

[Opalg]

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 $f(x) = e^x - \cos x - x$.