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I am just tryin to solve the limit:

$$\lim_{x\, \to \,\infty} \bigg(\frac{\sqrt[x]{1} + \sqrt[x]{2}}{2}\bigg)^x$$

(hope this isn't a duplicate, it is quite complicated to find special eq's via the search engine here) Wolfram-alpha told me it is $\sqrt{2}$. I have thought about using L'Hospital but the denominators derivatives to $f^{(n)} = (\ln2)^n \cdot 2^x \to \infty \; \forall n$.

So I don't see the use of it.

Hints are very welcome. (I already seen that $\sqrt[x]{1} = 1$ and tried the $e^{\ln f(x)}$ but without success)

Lu_kors
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4 Answers4

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Take $x= e^{\log x} $ and write $\frac{1}{x}=h$, so you have $\frac{\log (1+2^h)-\log 2 }{h}$, then use L'Hospital

Alex
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    Somewhat pedantic point: as is often the case, you've rewritten this as the definition of the derivative of $\ln \left ( 1 + 2^x \right )$ at $x=0$. So L'Hospital's rule is not necessary. Still nice, +1. – Ian Apr 22 '15 at 20:40
  • @Ian: This is exactly what I was trying to do (my favorite approach if you look at my similar answers), but for sm reasong couldn't get my head around making it precise, so jumped to L'H – Alex Apr 22 '15 at 20:41
  • Yup, works quite good; thank you. I haven't used L'Hospital here, chain rule is known :) – Lu_kors Apr 22 '15 at 21:00
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We have $$ y = \lim_{x\, \to \,\infty} \bigg(\frac{\sqrt[x]{1} + \sqrt[x]{2}}{2}\bigg)^x = \lim_{x\, \to \,\infty} \bigg(\frac{1 + 2^{1/x}}{2}\bigg)^x $$ So, $$ \ln y = \lim_{x\, \to \,\infty} x\ln\bigg(\frac{1 + 2^{1/x}}{2}\bigg) = \lim_{x\, \to \,\infty} \ln\bigg(\frac{1 + 2^{1/x}}{2}\bigg)/(1/x) $$ L'Hopital gives you $$ \ln y = \lim_{x\, \to \,\infty} \bigg(\frac{2}{1 + 2^{1/x}}\bigg) \cdot (-1/x^2)[\ln(2)/2]2^{1/x}/(-1/x^2) = \\ \lim_{x\, \to \,\infty} \bigg(\frac{2[\ln(2)/2]2^{1/x}}{1 + 2^{1/x}}\bigg) = \\ 2[\ln(2)/2] \lim_{x\, \to \,\infty} \bigg(\frac{2^{1/x}}{1 + 2^{1/x}}\bigg) =\\ 2[\ln(2)/2] \frac 12 = \ln(2)/2 $$ So, we have $y = e^{\ln(2)/2} = \sqrt 2$. So the limit is $\sqrt 2$.

Ben Grossmann
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Take the natural log of the argument of the limit:

$$ \ln\Bigl(\frac{1+\sqrt[x]{2}}{2}\Bigr)^x = x[\ln(1+\sqrt[x]{2})-\ln 2] $$

Now, in the limit, $[1+(\ln 2)/x]^x$ goes as $e^{\ln 2} = 2$, so $\sqrt[x]{2}$ goes as $1+(\ln 2)/x$. So the log of our argument now goes as $x[\ln(2+(\ln 2)/x)-\ln 2]$.

Observe that $d/du (\ln u) = 1/u = 1/2$ at $u = 2$, so the above expression goes as $x[\ln 2+(\ln 2)/(2x)-\ln 2] = (\ln 2)/2$. Since that is the log of our expression, the desired limit is $e$ raised to that power, or $\sqrt{2}$.

Brian Tung
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I think we can do this. Let $T(x)=a^x$. Then $T'(x)=a^x \ln(a)$ and so $T(0)=1$ and $T'(0)=\ln(a)$. $T'(0)=\ln(a)=\lim_{ u \rightarrow 0 } \frac{T(u)-T(0)}{u-0}=\lim_{ u \rightarrow 0 } \frac{a^u-1}{u} $ Let $u=\frac{1}{x}$ . This means we have: $\ln(a)=\lim_{ u \rightarrow \infty } (a^\frac{1}{x}-1)x$ So let's go back and play around with our problem: $\lim_{x \to \infty} (\frac{\sqrt[x]{1} + \sqrt[x]{2}}{2})^x=\lim_{x \to \infty}(\frac{(2^\frac{1}{x}-1)x}{2x}+1)^{2x \cdot \frac{1}{2}}=\lim_{x \to \infty} ((\frac{\ln(2)}{2x}+1)^{2x})^\frac{1}{2}=(e^{\ln(2)})^\frac{1}{2}$ Now if we can do this you might have to beef up some of what I have said. I will add for some small $\epsilon>0$ and large $x$ we have $\ln(a)=(a^\frac{1}{x}-1)x+\epsilon$ And so $\ln(a) \approx (a^\frac{1}{x}-1)x$

randomgirl
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