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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?

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  • $\begingroup$ Did you notice that you can simplify the first half as $\log(a/b)$, which is not indeterminate? $\endgroup$ – user296602 Aug 29 '17 at 21:07
  • $\begingroup$ @user296602 stupid me. And to think I wrestle with this for few hours. What should I do with this question when the answer came to my mind right away? $\endgroup$ – Hosea Aug 29 '17 at 21:09
  • $\begingroup$ You could delete it, or write your own answer for it. $\endgroup$ – user296602 Aug 29 '17 at 21:09
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$$\frac{\log(\frac{ax}{bx})}{\log\left(\dfrac{ax+(1-x)+c}{bx+(1-x)+c}\right)}$$

$$\frac{\log(\frac{a}b)}{\log\left(\dfrac{ax+(1-x)+c}{bx+(1-x)+c}\right)}$$

$$\frac{\log(\frac{a}b)}{\log\left(\dfrac{bx+(1-x)+c}{bx+(1-x)+c}+\dfrac{ax-bx}{bx+(1-x)+c}\right)}$$

$$\frac{\log(\frac{a}b)}{\log\left(1+\dfrac{x(a-b)}{bx+(1-x)+c}\right)}$$

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.

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  • $\begingroup$ Use \dfrac instead of \frac and use \left and \right when brackets are small, eg: $\left(\dfrac{a}{b}\right)$ \left(\drac{a}{b}}\right). $\endgroup$ – A---B Aug 29 '17 at 22:10
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Thanks to @user296602, I realized that the numerator results in a constant value log(a/b). Since denominator is 0, the limit will be +$\infty$ at $x = 0$.

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    $\begingroup$ How do you know the limit isn't $-\infty$? $\endgroup$ – Barry Cipra Aug 29 '17 at 21:27
  • $\begingroup$ @BarryCipra You are right, though I have a suspicion that a > b in my problem. It's part of bigger problem that I'm trying to solve, so I'll have to see if a > b indeed. $\endgroup$ – Hosea Aug 30 '17 at 13:13
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    $\begingroup$ It actually works either way: If $a\gt b$ then numerator and denominator are both positive; if $a\lt b$ then both are negative. (This assumes, of course, that $a$ and $b$ are both positive. Presumably this is true of $c$ as well.) $\endgroup$ – Barry Cipra Aug 30 '17 at 14:34

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