$\lim_{x\to 0} \left(\int_0^1 3y+2(1-y)^xdy\right)^{1/x}$ $$\lim_{x\to 0} \left(\int_0^1 \left(3y+2(1-y)^x\right)  dy\right)^{1/x}$$
I was solving this problem, and when I solve I got the answer which the software says incorrect.
My try:
Firstly, the simplest approach I directly solved the integration to get
$$\lim_{x\to 0} \left(\frac{3x+7}{2x+2}\right)^{\frac1x}=3.5^{\pm \infty}$$ not unique, so i thought it should be $\color{blue}{\text{Does not Exist}}$
Next approach was to use the shortcut formula of $1^\infty$ form involving exponent directly which gives me answer as $\frac1e$ which is also incorrect but later I feel it should be incorrect because when x tends to zero, the inner integral is not tends to 1.

Furthermore, if put the limit at the very start, I wanna know out of those 7 indeterminate forms $\color{red}{\text{Which form is this limit ?}}$

Do you also think the question is wrong ?

I think there is no limit. Solving the integral you have
\begin{eqnarray*}
    \int_{0}^{1} 3y+2(1-y)^{x}dy 
    & = &    \int_{0}^{1} 3ydy+\int_{0}^{1}2(1-y)^{x}dy\\
    & = &    \left.\left(\frac{3y^{2}}{2}-2\frac{(1−y)^{x+1}}{x+1}\right) 
\right|_{0}^{1}\\
    & = &    \frac{3}{2} + \frac{2}{x+1}
\end{eqnarray*}
Then we calculate the limit with $x=1/y$ a change of variable,
\begin{eqnarray*}
    \lim_{x\to0^{\pm}}\left(\frac{3}{2} + \frac{2}{x+1}\right)^{1/x}
    & = &    \lim_{x\to0^{\pm}}\left(\frac{3x+7}{2x+2}\right)^{1/x}\\
    & = &    \lim_{y\to\pm\infty}\left(\frac{\frac{3}{y}+7}{\frac{2}{y}+2}\right)^{y}\\
    & = &    \left(\frac{7}{2}\right)^{\pm\infty}
\end{eqnarray*}
this tiends to $0$ or $\infty$ depending on the sign, i.e. $(7/2)^{x}\to \infty$ when $x\to\infty$ and $(7/2)^{x}\to0$ when $x\to-\infty$
