Calculate $\lim_{x\, \to \,\infty} \bigg(\frac{\sqrt[x]{1} + \sqrt[x]{2}}{2}\bigg)^x$ 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)  
 A: 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
A: 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 $\sqrt 2$.
A: 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 $\sqrt{2}$.
A: 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 $\ln(a) \approx (a^\frac{1}{x}-1)x$
