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Compute $$ \lim_{x\to0}\left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2} $$ where $a$ and $b$ are positive numbers.

I came to the two different forms of this limit, as $\lim_{x\to0}$ enter image description here

$$e^{\frac{\ln b-\ln b}{\ln a-\ln a+\ln b-\ln b}}$$ $$\frac{x^2\cdot a^x+1-\frac{a^2}x}{x^2\cdot b^x+1-\frac{b^2}x}$$

So, what I want to say is that I can't solve this problem and I'm here for any kind of help.

Martin Sleziak
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Музаффар Шакаров
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    Expanding $$c^x=e^{x\log c}=1+x\log c+\tfrac12x^2(\log c)^2+o(x^2)$$ for $c=a$ and for $c=b$ yields directly the limit $$\exp(\tfrac12((\log a)^2-(\log b)^2))$$ Of course, no L'Hôpital is needed. – Did Jan 09 '17 at 11:49
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    For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Jan 09 '17 at 13:06

2 Answers2

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Compute the limit of the logarithm of your function $$ \lim_{x\to0}\ln\left( \left(\frac{a^x-x\ln a}{b^x-x\ln b}\right)^{1/x^2} \right)= \lim_{x\to0} \frac{\ln(a^x-x\ln a)-\ln(b^x-x\ln b)}{x^2} $$ under the assumption that $a\ne b$ (if $a=b$ the limit is obvious).

Here applying L’Hôpital is quite easy. If you find the limit is $l$, then the required limit is $e^l$.

Note that you can just compute $$ l_a=\lim_{x\to0}\frac{\ln(a^x-x\ln a)}{x^2} $$ because the given limit will then be $l_a-l_b$.

egreg
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Assuming that the desired limit is $L$ we can proceed as follows: \begin{align} \log L &= \log\left\{\lim_{x \to 0}\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\right\}\notag\\ &= \lim_{x \to 0}\log\left(\frac{a^{x} - x\log a}{b^{x} - x\log b}\right)^{1/x^{2}}\text{ (via continuity of log)}\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\log\left(1 + \frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\cdot\dfrac{\log\left(1 + \dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\right)}{\dfrac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}}\notag\\ &= \lim_{x \to 0}\frac{1}{x^{2}}\cdot\frac{a^{x} - x\log a - b^{x} + x\log b}{b^{x} - x\log b}\notag\\ &= \lim_{x \to 0}\frac{a^{x} - b^{x} - x\log a + x\log b}{x^{2}}\tag{1}\\ &= \lim_{x \to 0}\frac{a^{x}\log a - b^{x}\log b - \log a + \log b}{2x}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\log a\cdot\frac{a^{x} - 1}{x} - \log b\cdot\frac{b^{x} - 1}{x}\notag\\ &= \frac{(\log a)^{2} - (\log b)^{2}}{2}\notag \end{align} If you prefer the easier technique of Taylor series then you need to apply it after step marked $(1)$ and use the expansion $$a^{x} = 1 + x\log a + \frac{x^{2}}{2}(\log a)^{2} + o(x^{2})$$ This way you get the answer immediately.

Thus finally $$L = \exp\left(\frac{(\log a)^{2} - (\log b)^{2}}{2}\right)$$

Paramanand Singh
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