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

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

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 did that and got 1/e as answer which was also wrong Commented May 19, 2021 at 17:16
• @fGDu94 you cannot apply l'hopital on the exponent because the numerator does not go to $0$ or $\pm \infty$. Commented May 19, 2021 at 17:25
• exactly, we dont have here 0/0 or inf/inf form Commented May 19, 2021 at 17:25
• Your try is correct. The limit indeed does not exist. Commented May 19, 2021 at 17:29
• Are you sure you got the problem right? I've got the same answer as your first approach Commented May 19, 2021 at 17:30

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 $$x\to-\infty$$
• what did you do to $e^{x \ln(1-y)}$? Commented May 19, 2021 at 17:34
• You’ve also lost the $2.$ It should be $$\frac32+\frac2{x+1}=\frac{3x+7}{2x+2}$$ which just reproduces what OP got. Commented May 19, 2021 at 18:22