What is $\lim_{p \to 0} \left(\int_0^1 (1+x)^p \, dx\right)^{1/p}$? What is $$\lim_{p \to 0} \left(\int_0^1 (1+x)^p \, dx\right)^{1/p}\text{ ?}$$ 
I used binomial to get value as $2^p-1$ so limit becomes $$(2^p-1)^{1/p}.$$ But I can't go any further. 
 A: We have that
$$\int_0^1(1+x)^pdx=\begin{cases}\left.\log(1+x)\right|_0^1=\log2\;,&p=-1\\{}\\\left.\frac{(1+x)^{p+1}}{p+1}\right|_0^1=\frac1{p+1}\left(2^{p+1}-1\right)\;:&p\neq-1\end{cases}$$
so as $\;p\to 0\;$ we can safely assume the second option above, and then
$$\left(\int_0^1(1+x)^pdx\right)^{1/p}=\left(\frac{2^{p+1}-1}{p+1}\right)^{1/p}=$$
$$=\left[\frac1{(1+p)^{1/p}}\right](2^{p+1}-1)^{1/p}\xrightarrow[p\to0]{}\frac1e\cdot4=\frac4e$$
since
$$\lim_{p\to0}\log\left(2^{p+1}-1\right)^{1/p}=\lim_{p\to0}\frac{\log(2^{p+1}-1)}p\stackrel{\text{l'Hospital}}=$$
$$=\lim_{p\to0}\frac{2^{p+1}\log2}{2^{p+1}-1}=\frac{2\log2}1\implies\lim_{p\to0}(2^{p+1}+1)=e^{2\log2}=4$$
A: Apply $\ln $ to the expression to get
$$\frac{\ln [\int_0^1 (1+x)^p\, dx]}{p}.$$
Now as $p\to 0,$ the integral inside the brackets $\to 1,$ simply because $(1+x)^p \to 1$ uniformly. So we have a $0/0$ sitution and can apply L'Hopital. The quotient of derivatives is
$$\tag 1 \frac{\int_0^1 \ln (1+x)(1+x)^p\, dx}{\int_0^1 (1+x)^p\, dx}.$$
Here I've used "differentiation under the integral sign", which is simple to verify here. Again use $(1+x)^p \to 1$ uniformly to see the limit of $(1)$ is $\int_0^1 \ln (1+x), dx.$ That integral equals $\ln (4/e).$  Exponentiating back gives $4/e$ for the original limit.
A: $$X = \lim_{p\to 0} (2^p-1)^{1/p}$$
$$\log X = \lim_{p\to 0} 1/p \log(2^p-1) = -\infty$$
$$X=0$$
A: $$\lim_{p\to0}\int^{1}_{0} y^p dy$$
$$\lim_{p\to0} (y^{p+1}/p+1|_{1}^{2})^{1/p}$$
it reduces to $$\lim_{p\to0}[2^{p+1}/{p+1}-1/{p+1}]^{1/p}$$
now by applying limit 
$$\log X=\lim_{p\to0}1/p \log [2^{p+1}/{p+1}-1/{p+1}]$$
apply L hospitals 
$$\frac{\frac{\log (2)(2^{p+1})(p+1)-(2^{p+1}-1)}{(p+1)^2}}{[2^{p+1}/{p+1}-1/{p+1}]}$$
now putting the $\lim p \to 0$
$$\ln X =\ln(4/e)$$
$$X=4/e$$
A: Consider
$$
\int_0^1(1+x)^p\,dx=\left[\frac{(1+x)^{p+1}}{p+1}\right]_0^1=
\frac{2^p-1}{p+1}
$$
(we may consider $p>-1$, of course). Now you want
$$
\lim_{p\to0}\left(\frac{2^{p+1}-1}{p+1}\right)^{\!1/p}
$$
and it's better to compute the limit of the log:
$$
\lim_{p\to0}\frac{\log(2^{p+1}-1)-\log(p+1)}{p}
$$
You can do it separately:
$$
\lim_{p\to0}\frac{\log(2^{p+1}-1)}{p}
$$
is the derivative at $0$ of $f(p)=\log(2^{p+1}-1)$; since
$$
f'(p)=\frac{1}{2^{p+1}-1}2^{p+1}\log2
$$
we have $f'(0)=2\log2$.
On the other hand, it should be well known that
$$
\lim_{p\to0}\frac{\log(p+1)}{p}=1
$$
So your limit is
$$
2\log2-1=\log\frac{4}{e}
$$
and your original limit is $4/e$.
A: In the same spirit as egreg's answer, let us consider $$A_p=\left(\frac{2^{p+1}-1}{p+1}\right)^{\!1/p}$$ Take logarithms $$\log(A_p)=\frac 1p \log\left(\frac{2^{p+1}-1}{p+1}\right)$$ Now, from definition, build Taylor series around $p=0$ to get $$\log\left(\frac{2^{p+1}-1}{p+1}\right)=p (2 \log (2)-1)+p^2 \left(\frac{1}{2}-\log ^2(2)\right)+O\left(p^3\right)$$ So $$\log(A_p)=(2 \log (2)-1)+p \left(\frac{1}{2}-\log ^2(2)\right)+O\left(p^2\right)$$ Now, using $A_p=e^{\log(A_p)}$ and  Taylor series again $$A_p=\frac{4}{e}+\frac{2 -4\log ^2(2)}{e}p+O\left(p^2\right)$$ which shows the limit and also how it is approached.
For illustration purposes, let us try using $p=\frac 1{10}$. The exact result is $$A_{\frac 1 {10}}=\Big(\frac{10}{11} \left(2 \sqrt[10]{2}-1\right)\Big)^{10}\approx 1.474392296$$ while the approximation would give $$\frac{21-2 \log ^2(2)}{5 e}\approx 1.474394138$$ Using one more term in Taylor expansions would lead to an an approximate value $\approx 1.474392407$.
