$$\lim_{x\to 0} \left(\int_0^1 \left(3y+2(1-y)^x\right) dy\right)^{1/x}$$
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 $x\to-\infty$