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A friend asked me to help him with calculating a certain hideous limit: $$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x},\space\space0<a,b,c\in\mathbb{R}$$

I came up with a solution (and wolfram alpha confirmed), but unfortunately it involves a lot of steps and a couple of theorems and identities so we're actually reaching out hoping someone can come up with a better solution. Hopefully a more intuitive one!

Here's what I had in mind:

$$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\underset{x\rightarrow0}{\lim}e^{\ln{\big((a^{x}+b^{x}-c^{x})^\frac{1}{x}\big)}}=\underset{x\rightarrow0}{\lim}e^\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}$$

Since $e^x$ is continuous and therefore we have: $$\underset{x\rightarrow{x_0}}{\lim}e^{f(x)}=e^{\underset{x\rightarrow{x_0}}{\lim}f(x)}$$

So we focus on finding: $$\underset{x\rightarrow0}{\lim}\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}$$ Suppose it is equal to some $L$, then our original limit will be equal to $e^L$.

We now note that: $$\underset{x\rightarrow0}{\lim}x=0\space,\space \underset{x\rightarrow0}{\lim}\ln{\big(a^{x}+b^{x}-c^{x}\big)}=\ln{(a^0+b^0-c^0)}=\ln{(1+1-1)}=\ln(1)=0$$

So our limit takes the indeterminate form $"\frac{0}{0}"$. After checking all the conditions are satisfied we proceed with L'Hospital setting $g(x)=x$ and $f(x)=\ln{\big(a^{x}+b^{x}-c^{x}\big)}$, and get the following: (oh boy this is going to be an ugly)

$$\underset{x\rightarrow0}{\lim}\frac{f(x)}{g(x)}=\underset{x\rightarrow0}{\lim}\frac{f'(x)}{g'(x)}=\underset{x\rightarrow0}{\lim}\frac{\frac{a^{x}\ln{(a)}+b^{x}\ln{(b)}-c^{x}\ln{(c)}}{a^{x}+b^{x}-c^{x}}}{1}=\ln{(a)}+\ln{(b)}-\ln{(c )}=\ln{\big(\frac{ab}{c}\big)}$$

(Again, the computation was straightforward because $\ln$ is continuous and the limit of composition of functions).

So we finally get: $$\underset{x\rightarrow0}{\lim}\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\underset{x\rightarrow0}{\lim}e^\frac{\ln{\big(a^{x}+b^{x}-c^{x}\big)}}{x}=e^{\ln{(\frac{ab}{c})}}=\frac{ab}{c}$$

And that's it! As you can see it's not that intuitive, and it involves a lot of computation and some theorems and identities and as a result- many steps. I would appreciate in other insight regrading formality and other perspectives on calculating this limit. Thank you

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By substituting $x=1/n$ and taking the limit from the right we get

$$\lim_{n\to\infty}\left(a^{1/n}+b^{1/n}-c^{1/n}\right)^n$$

Applying Maclaurin series for $a^x=1+x\ln(a)+\mathcal O(x^2)$ we get:

$$a^{1/n}+b^{1/n}-c^{1/n}=1+\frac1n\ln\left(\frac{ab}c\right)+\mathcal O\left(\frac1{n^2}\right)$$

Recall the alternative definition $e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$ and you can show that we hence have:

$$\lim_{n\to\infty}\left(a^{1/n}+b^{1/n}-c^{1/n}\right)^n=e^{\ln(ab/c)}=\frac{ab}c$$


More formal arguments can be made to show that $e^x=\lim\limits_{n\to-\infty}\left(1+\frac xn\right)^n$ and hence $e^x=\lim\limits_{t\to0}\sqrt[t]{1+xt}$, and furthermore that $e^x=\lim\limits_{t\to0}\sqrt[t]{1+xt+o(t)}$.

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  • $\begingroup$ Thank you very much! My friend is taking calc1 now so she didn't study Maclaurin\Taylor series yet. But I couldn't figure out how to do that using Maclaurin\Taylor series and your solution is really great and insightful! Much appreciated! $\endgroup$ – omer Sep 18 '19 at 22:03
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More generally, if $f(x) =\left(\sum_{k=1}^n c_ia_i^x\right)^{1/x} $, then $f(x) =e^{g(x)} $ where $g(x) =\ln\left(\sum_{k=1}^n c_ia_i^x\right)^{1/x} =\frac1{x}\ln\left(\sum_{k=1}^n c_ia_i^x\right) $.

Then, as $x \to 0$,

$\begin{array}\\ g(x) &=\frac1{x}\ln\left(\sum_{k=1}^n c_ia_i^x\right)\\ &=\frac1{x}\ln\left(\sum_{k=1}^n c_ie^{\ln(a_i)x}\right)\\ &=\frac1{x}\ln\left(\sum_{k=1}^n c_i(1+\ln(a_i)x+O(x^2))\right)\\ &=\frac1{x}\ln\left(\sum_{k=1}^n c_i+x\sum_{k=1}^n c_i\ln(a_i)+O(x^2))\right)\\ &=\frac1{x}\ln (C\left(1+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2)\right)) \qquad C=\sum_{k=1}^n c_i\\ &=\frac1{x}(\ln (C)+\ln\left(1+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2)\right))\\ &=\frac1{x}(\ln (C)+\frac{x}{C}\sum_{k=1}^n c_i\ln(a_i)+O(x^2))\\ &=\frac{\ln(C)}{x}+\frac{1}{C} \ln(\prod_{k=1}^na_i^{c_i})+O(x)\\ \text{if }C=1 &\text{as in this case }(C=1+1-1=1)\\ g(x) &= \ln(\prod_{k=1}^na_i^{c_i})+O(x)\\ \text{so}\\ f(x) &e^{g(x)}\\ &=\prod_{k=1}^na_i^{c_i}+O(x)\\ \text{if } C > 1\\ g(x) &\to \infty\\ \text{so}\\ f(x) &\to \infty\\ \text{if } C < 1\\ g(x) &\to -\infty\\ \text{so}\\ f(x) &\to 0\\ \end{array} $

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$$y=\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}\implies \log(y)=\frac{1}{x}\log\big(a^{x}+b^{x}-c^{x}\big)$$ Now, using $t^x=e^{x \log(t)}$ and using Taylor expansion around $x=0$, we have $$t^x=1+x \log (t)+\frac{1}{2} x^2 \log ^2(t)+O\left(x^3\right)$$ Applying it for $t=a,b,c$,we then have $$a^{x}+b^{x}-c^{x}=1+x (\log (a)+\log (b)-\log (c))+\frac{1}{2} x^2 \left(\log ^2(a)+\log ^2(b)-\log ^2(c)\right)+O\left(x^3\right)$$ Let $$A=(\log (a)+\log (b)-\log (c))\qquad\text{and} \qquad B=\frac{1}{2} \left(\log ^2(a)+\log ^2(b)-\log ^2(c)\right)$$ to make $$a^{x}+b^{x}-c^{x}=1+A x+ B x^2$$ Continuing with Taylor $$\log(1+A x+ B x^2)=A x+ \left(B-\frac{A^2}{2}\right)x^2+O\left(x^3\right)$$ $$\log(y)=A + \left(B-\frac{A^2}{2}\right)x+O\left(x^2\right)$$ which, for sure, shows the limit but also how it is approched. Back to the definition of $A,B$ and simplifying, then, close to $x=0$ $$\big(a^{x}+b^{x}-c^{x}\big)^\frac{1}{x}=\log \left(\frac{a b}{c}\right)+\log \left(\frac{a}{c}\right) \log \left(\frac{c}{b}\right)x+O\left(x^2\right)$$

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