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I need to solve the limit $$\lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}$$ when $a,b,c \gt 0$. I'm looking for ways to avoid $\frac{1}{x}$ power.

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I formatted your question. Please make certain I have done so faithfully. In particular, you mention "sqrt of $x$" in your question, but it did not appear in the limit you wrote down. Are you referring to the "$\frac{1}{x}$ power"? –  Austin Mohr Feb 21 '12 at 18:38
    
Yes Austin that is correct thank you. –  Tom Feb 21 '12 at 18:39
    
The answer (if I am not wrong) is $\frac{ab}{c}$. I'll try to punch in a solution soon. –  Inquest Feb 21 '12 at 18:49
    
@Nunoxic I went to my board, did the calculations and got the same. My comment'd have been the same as yours. –  Pedro Tamaroff Feb 21 '12 at 21:24

5 Answers 5

up vote 6 down vote accepted

Note that $$ \lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x} =\lim_{x \rightarrow 0}\,e^{\left(\displaystyle\frac1x \log(a^x+b^x-c^x)\right)}. $$ So we only need to calculate $$ \lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x). $$ Since the expression inside the log goes to $1$ as $x\to0$, we can apply L'Hôpital, to get $$ \lim_{x \rightarrow 0}\frac1x \log(a^x+b^x-c^x) =\lim_{x\to0}\frac{a^x\,\log a+ b^x\,\log b-c^x\,\log c}{a^x+b^x-c^x} =\log a+ \log b-\log c. $$ Then $$ \lim_{x \rightarrow 0} (a^x + b^x - c^x)^\frac{1}{x}=e^{\log a+ \log b-\log c}=\frac{ab}c $$

Added: as suggested by Aryabhata, it remains to justify that one can actually apply L'Hôpital. For the rule to apply, we need to have a quotient $f(x)/g(x)$, with both $f$ and $g$ differentiable at $0$, with $g'(x)\ne0$ on a neighbourhood of $0$, such that $f(0)=g(0)=0$ (less is needed, in fact, only that they go to $0$) and such that $\lim_{x\to0}f'(x)/g'(x)$ exists (it equals $ab/c$ in this case). Here $f(x)=\log(a^x+b^x-c^x)$, $g(x)=x$. The function $x\mapsto a^x+b^x-c^x$ is differentiable everywhere; for $x$ near $0$, the values of this function approximate $1$; on any neighbourhood of $1$ that doesn't contain $0$, $\log$ is differentiable, and then so is the composition $f(x)=\log(a^x+b^x-c^x)$. And $g'(x)=1\ne0$ for all $x$.

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This seems circular: LHospital assumes that the derivative exists at $0$. But the limit you are trying to find is exactly the derivative at $0$! (This is what I was hinting with my answer). –  Aryabhata Feb 21 '12 at 19:43
    
@Aryabhata: I'm using L'Hopital to find the value of the limit, not to prove that the derivative exists. The function $\log(a^x+b^x-c^x)$ is C$^\infty$ on small enough neighbourhoods of $0$. –  Martin Argerami Feb 21 '12 at 20:59
    
Finding a limit also entails proving its existence (implicitly or explicitly). Otherwise, all you have done is, "if the limit exists, then it is so and so", which is incomplete. –  Aryabhata Feb 21 '12 at 21:08
    
@Aryabhata: L'Hôpital is exactly about the existence of the limit. –  Martin Argerami Feb 21 '12 at 21:19
    
You missed mentioning that $g'(x) \ne 0$ in a neighbourhood of $0$, but that is fine here as $g(x) = x$. Sure, Lhopital also proves existence, but that was not my point. –  Aryabhata Feb 21 '12 at 21:37

Another hint: Let $f(x) = \log (a^x + b^x - c^x)$

What is $f(0)$?

What is $f'(0)$? (The derivative at $0$).

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f(0) is 0 and f(tag)(0) is minus infinity..trying to think what this tells me –  Tom Feb 21 '12 at 19:01
    
@TOm: What is the definition of $f'(0)$ in terms of limits? –  Aryabhata Feb 21 '12 at 19:05
    
@Tom: Are you sure about $f'(0)$? –  Aryabhata Feb 21 '12 at 19:11

HINT: $$ \left( a^x + b^x - c^x \right)^\frac{1}{x} = a \left( 1 + \left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x \right)^\frac{1}{x} = a \left( 1 + x \cdot \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x} \right)^\frac{1}{x} $$ Compute $\lim_{x \to 0} \frac{\left(\frac{b}{a}\right)^x - \left(\frac{c}{a}\right)^x}{x}$, then use the golden limit $\lim_{x\to 0} \left( 1 + \alpha x\right)^\frac{1}{x}$.

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This needs some justification. For instance consider limit of $(1+x)^{1/x}$ as $x \to 0$ (which is $e$). We just cannot replace $(1+x)$ by $1$ because $x \to 0$. –  Aryabhata Feb 21 '12 at 18:54
    
@Aryabhata Sasha is not suggesting that. –  Pedro Tamaroff Feb 21 '12 at 19:01
    
this looks like magic but yet again the limit to compute when x approaches 0 is hard for me to do –  Tom Feb 21 '12 at 19:05
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@Peter: I am pretty sure Sasha is not doing that, but people who are just learning these things might think that that is what Sasha is doing. Just pointing out the potential pitfall. –  Aryabhata Feb 21 '12 at 19:43

Where you start is to realize that $1^\infty$ is an indeterminate form, so you can use L'hopital's rule after you rewrite it as,

$$e^{\left(\lim\limits_{x \rightarrow 0} \frac{\ln(a^x + b^x - c^x)}{x}\right)}$$

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If you want to avoid L'Hopital: Write $(a^x + b^x - c^x)^{1 \over x}$ as $a(1 + ({b \over a})^x - ({c \over a})^x)^{1 \over x}$, so that it suffices to find $$\lim_{x \rightarrow 0} (1 +({b \over a})^x - ({c \over a})^x)^{1 \over x}$$ Taking logs, it's enough to find $$\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) \over x}$$ $$=\lim_{x \rightarrow 0} {\ln(1 +({b \over a})^x - ({c \over a})^x) - \ln(1 +({b \over a})^0 - ({c \over a})^0) \over x - 0}$$ This is the difference quotient for the derivative of $\ln(1 +({b \over a})^x - ({c \over a})^x)$ at $x = 0$. Using the chain rule to find this derivative and plugging in $x = 0$ gives you $\ln(b/a) - \ln(c/a) = \ln(b/c)$. Going backwards, the original limit is ${\displaystyle ae^{\ln(b/c)} = {ab \over c}}$.

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