# I need a help with this limit?

I try to understand what magic trick I need to prove that

$$\lim_{x\rightarrow \infty} \left(\frac{a^{1/x}+b^{1/x}}{2}\right)^x = \sqrt{ab}.$$

If I evaluate directly, I get indetermination $1^{\infty}$, but i tried with the usual hint and i can't do that. Any suggestions? Thanks a lot.

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Let's give a simple proof based on the Taylor series although a more general and nice answer is given by Daniel Fischer. Recall that $$e^u\sim_0 1+u\quad;\quad\ln(1+u)\sim_0 u$$ so $$\frac{a^{1/x}+b^{1/x}}{2}\sim_{\infty}1+\frac1{2x}(\ln a+\ln b)=1+\frac1x\ln(\sqrt{ab})$$ hence $$\left(\frac{a^{1/x}+b^{1/x}}{2}\right)^x\sim_\infty\exp \left(x\log\left(1+\frac1x\ln(\sqrt{ab})\right)\right)\sim_\infty\sqrt{ab}$$

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Use List of indeterminate forms. So, whenever you have $\lim_{x \to c} f(x) = 1$ and $\lim_{x \to c} g(x) = \infty$, then you can do $$\lim_{x \to c} f(x)^{g(x)} = \exp \left(\lim_{x \to c} \frac {\ln f(x)}{\frac 1{g(x)}} \right )$$ which has the form of $\frac 00$, sou can apply L'Hôpital's rule $$\lim_{x \to c} f(x)^{g(x)} = \exp \left( \lim_{x \to c} \frac {\left( \ln f(x)\right)'}{\left( \frac 1{g(x)}\right)'}\right ) = \exp \left( -\lim_{x \to c} \frac {f'(x) g^2(x)}{f(x) g'(x)}\right)$$ In you case $$f(x) = \frac {a^{\frac 1x} + b^{\frac 1x}}2, \quad g(x) = x, \quad c = \infty$$ and $$f'(x) = -\frac {\ln a \cdot a^{\frac 1x} + \ln b \cdot b^{\frac 1x}}2 \cdot \frac 1{x^2}, \quad g'(x) = 1$$ so $$\lim_{x \to c} f(x)^{g(x)} = \exp \left( \lim_{x \to \infty} \frac {\ln a \cdot a^{\frac 1x} + \ln b \cdot b^{\frac 1x}}{a^{\frac 1x} + b^{\frac 1x}}\right ) = \exp \left( {\frac {\ln a + \ln b}2} \right ) = \exp \left( \frac {\ln ab}2 \right ) = \sqrt{ab}$$

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In a comment Daniel Fischer has linked to a more general version. However, there may be some interest in having a detailed evaluation of the present simpler case available at math StackExchange.

It will be more convenient later on (when taking derivatives for L'Hopital's rule) to make the variable change $u = \frac{1}{x},$ which gives rise to the following equivalent limit: $$L \;\; = \;\; \lim_{u \rightarrow 0} \left( \frac{a^u + b^u}{2} \right)^{\frac{1}{u}}$$

Note: Actually, the correct counterpart to "$x \rightarrow \infty$" is $u \rightarrow 0^{+}$" $(u$ approaches $0$ from the right). However, for notational simplicity and because both unilateral limits exist as $u$ approaches $0,$ I'm going to write $u \rightarrow 0$ in what follows.

Take the logarithm of both sides: $$\ln L \;\; = \;\; \ln \left[ \lim_{u \rightarrow 0} \left( \frac{a^u + b^u}{2} \right)^{\frac{1}{u}} \right]$$ Use continuity of the logarithm function: $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[ \ln \left( \frac{a^u + b^u}{2} \right)^{\frac{1}{u}} \right]$$ Rewrite using an algebraic logarithm property: $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[ \frac{1}{u} \cdot \ln \left( \frac{a^u + b^u}{2} \right) \right]$$ Use the fact that multiplication by $\frac{1}{u}$ is equivalent to division by $u$: $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[ \frac{\ln \left( \frac{a^u + b^u}{2} \right)}{u} \right]$$ Rewrite using an algebraic logarithm property: $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[\frac{\ln ( a^u + b^u )\;-\;\ln 2}{u}\right]$$ Replace with the "L'Hopital rule differentiated version": $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[\frac{\frac{a^u\ln a \; + \; b^u \ln b}{a^u \; + \; b^u} \; - \; 0}{1} \right]$$ $$\ln L \;\; = \;\; \lim_{u \rightarrow 0} \left[ \frac{a^u\ln a \; + \; b^u \ln b}{a^u \; + \; b^u} \right]$$ Evaluate the limit: $$\ln L \;\; = \;\; \left[ \frac{a^0\ln a \; + \; b^0 \ln b}{a^0 \; + \; b^0} \right]$$ $$\ln L \;\; = \;\; \left[ \frac{\ln a \; + \; \ln b}{2} \right]$$ Solve for $L$ and rewrite until we get the desired result: $$L \;\; = \;\; e^{\frac{\ln a \; + \; \ln b}{2}}$$ $$L \;\; = \;\; e^{(\ln a \; + \; \ln b) \cdot \frac{1}{2}}$$ $$L \;\; = \;\; \left[ e^{(\ln a \; + \; \ln b)} \right]^{\frac{1}{2}}$$ $$L \;\; = \;\; \left[ e^{\ln a} \cdot e^{\ln b} \right]^{\frac{1}{2}}$$ $$L \;\; = \;\; \left[ab \right]^{\frac{1}{2}}$$ $$L \;\; = \;\; \sqrt{ab}$$

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O-K, I've gotten two down votes for this after two days, and I can't figure out what the shortcoming might be. I tried to make this as clear and as explicit as I could using only ideas and methods from a first year calculus course. If I made an error somewhere, or if I skipped/assumed something that is not line with this goal (a goal that should be obvious to more advanced readers), I sure don't see it. – Dave L. Renfro Jun 6 '14 at 13:38