This is a question from a mock exam of national engineering test in my country.
Firstly, we can't take limit inside the integral as the limits of integral are not independent.
I also tried applying the property:
$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
but that also doesn't give anything significant which can help in simplifying the problem further.
Leibniz integral rule also doesn't seem to give anything useful. I would be grateful if I could get more ideas from the community on how to approach this.
Best Answer
Actually you can exchange $\lim$ and $\int$.
The sequence of non-negative functions
$$ g_k(x) = \left\{\begin{array}{rcl}\left(1-\frac{x}{k}\right)^k&\text{if}&0\leq x\leq k\\0&\text{if}&x\geq k\end{array}\right. $$ converges monotonically to $e^{-x}$. By the dominated/monotone convergence theorem it follows that
$$ \lim_{k\to +\infty}\int_{0}^{+\infty} g_k(x) e^{x/3}\,dx =\int_{0}^{+\infty}e^{-2x/3}\,dx=\color{red}{\frac{3}{2}}.$$
Provided that $g_k(x)\leq e^{-x}$, we can also study how fast the sequence converges to $\frac{3}{2}$. Indeed
$$ \int_{0}^{+\infty}(e^{-x}-g_k(x))e^{x/3}\,dx = \int_{0}^{k}\left((e^{-x/k})^k-\left(1-\frac{x}{k}\right)^k\right)e^{x/3}\,dx + \int_{k}^{+\infty}e^{-2x/3}\,dx $$ where the last integral in the RHS equals $\frac{3}{2}e^{-2k/3}$ and the first one is $$ I_k=k\int_{0}^{1}\left(\left(e^{-x}\right)^k-(1-x)^k\right) e^{kx/3}\,dx. $$ Over $[0,1]$ we have $e^{-x}-(1-x)\leq \frac{1}{2}x^2$, and $a\geq b\geq 0$ implies $$ a^n-b^n = (a-b)(a^{n-1}+\ldots+b^{n-1}) \leq n(a-b)a^{n-1}, $$ so $I_k\geq 0$ is bounded by
$$ k^2\int_{0}^{1}\frac{x^2 e^x}{2}e^{-kx} e^{kx/3}\,dx\leq \frac{ek^2}{2}\int_{0}^{1}x^2 e^{kx/3-kx}\,dx\leq\frac{ek^2}{2}\int_{0}^{+\infty}x^2 e^{-2kx/3}\,dx $$ i.e. $$ I_k \leq \frac{ek^2}{2}\cdot\frac{27}{4k^3}=O\left(\frac{1}{k}\right). $$