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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.

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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. – 2017-01-09
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    For some basic information about writing math at this site see e.g. [here](/help/notation), [here](//meta.math.stackexchange.com/q/5020), [here](//meta.stackexchange.com/a/70559) and [here](//meta.math.stackexchange.com/q/1773). – 2017-01-09

2 Answers 2

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

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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 $l_a-l_b$.

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    How did u come to ur first limit? – 2017-01-09
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    @МузаффарШакаров I added the first step – 2017-01-09
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    +1 For that "note that ...". As both separate limits exist finitely one can do that, otherwise applying l'Hospital directly to the right hand side in the first line is an utter horror. – 2017-01-09