I'm trying to find the limit of the following function when $x = 0$.
$$\frac{\log\left(\dfrac{ax}{bx}\right)}{\log\left(\dfrac{ax + (1-x) + c}{bx + (1-x) + c}\right)}$$
where a,b and c is a constant.
It is my understanding that the at $x = 0$, numerator amounts to indeterminate ($\log(0) - \log(0)$). Denominator amounts to $0$ ($\log(1+c) - \log(1+c)$).
I recall that L'hospital rule is not applicable unless the term is in the form of $\frac{0}{0}$ or $\frac{\infty}{\infty}$. However, my numerator is indeterminate. I've nonetheless tried applying L'Hospital rule, but the numerator will just keep amounting to $\infty - \infty$ like above.
How can I find the limit of this function at 0?
$$\frac{\log(\frac{ax}{bx})}{\log\left(\dfrac{ax+(1-x)+c}{bx+(1-x)+c}\right)}$$
$$\frac{\log(\frac{a}b)}{\log\left(\dfrac{ax+(1-x)+c}{bx+(1-x)+c}\right)}$$
$$\frac{\log(\frac{a}b)}{\log\left(\dfrac{bx+(1-x)+c}{bx+(1-x)+c}+\dfrac{ax-bx}{bx+(1-x)+c}\right)}$$
$$\frac{\log(\frac{a}b)}{\log\left(1+\dfrac{x(a-b)}{bx+(1-x)+c}\right)}$$
x is infinitesimal, and I can assume you were using natural logarithms (though it wouldn't matter either way), so we can make the following generalization.
$$\log\left(1+\dfrac{x(a-b)}{bx+(1-x)+c}\right)= \log(1)+\dfrac{\dfrac{x(a-b)}{bx+(1-x)+c}}1$$
$$\log\left(1+\dfrac{x(a-b)}{bx+(1-x)+c}\right)=\dfrac{x(a-b)}{bx+(1-x)+c}$$
(since the derivative of ln(x) is $\frac{1}x$)
going back to our original problem,
$$\dfrac{\log(\frac{a}b)}{\log\left(1+\dfrac{x(a-b)}{bx+(1-x)+c}\right)}$$
$$\dfrac{\log(\frac{a}b)}{\dfrac{x(a-b)}{bx+(1-x)+c}}$$
Now, since all the numerators and denominators are cleared up, we can use 0 instead of x
$$\frac{\log(\dfrac{a}b)}{\dfrac{0(a-b)}{0\cdot b+(1-0)+c}}$$
$$\frac{\log(\dfrac{a}b)}{\dfrac{0(a-b)}{1+c}}$$
$$\frac{\log\left(\dfrac{a}b\right)(1+c)}{0(a-b)}$$
$$\infty \cdot \frac{\log\left(\dfrac{a}b\right)(1+c)}{a-b}$$
That's the essential answer. Technically, I could just write it as $\infty$, but it's more convenient to have a multiplier. If you want to know the rest of the series to the answer (as in $A\infty²+B\infty+C+D\Delta+E\Delta²+...$), just say so in the comments, as it's sometimes helpful to know the continued series if you're going to continue performing limits on your equation or if you want to represent a taylor's series to your function about x=0.