Evaluate the limit of an integral $$\lim_{y\to0} \int_{0}^{2} \frac{e^{(x+y)^2}-e^{x^2}}{y} dx$$
let $y_n\to 0$
define $f_n(x)=\frac{e^{(x+y_n)^2}-e^{x^2}}{y_n}$
it seems that we can use mean value theorem for $e^{x^2}$, and use DCT. Is my thought correct?
 A: We don't need to invoke the mean value theorem, or the Dominated Convergence Theorem, or both to evaluate the coveted limit.  
Rather we simply write the term of interest as
$$\int_0^2 \frac{e^{(x+y)^2}-e^{x^2}}{y}\,dx=\frac{\int_y^{y+2} e^{x^2}\,dx-\int_0^2 e^{x^2}\,dx }{y}$$
Then, applying the definition of the derivative of $f(y)=\int_y^{y+2}e^{x^2}\,dx$ at $y=0$ immediately gives
$$\lim_{y\to 0}\int_0^2 \frac{e^{(x+y)^2}-e^{x^2}}{y}\,dx=\lim_{y\to 0}\left(e^{(y+2)^2}-e^{y^2}\right)=e^4-1$$
And we are done!
A: If $f(x) = e^{x^2}$ then the mean value theorem tell us 
$$ g(y) = \frac{e^{(x+y)^2} - e^{x^2}}{y} = \frac{e^{(x+y)^2} - e^{x^2}}{(x+y) -x } = f_{y}'(c) $$ where the notation implies that for different values of $y$, you get different values of $f'$.  But for $x \in [0,2]$ we have that 
$f_{y}'(c) = g(y) < \max_{y \in [0,2]}\{f'(y)\} =  f'(2)$ so we may apply DCT and get 
$$ \lim_{y \to 0}\int_{0}^{2} \frac{e^{(x+y)^2} - e^{x^2}}{y} dx = \int_{0}^{2} \lim_{y \to 0} \frac{e^{(x+y)^2} - e^{x^2}}{y} dx = \int_{0}^{2} 2xe^{x^2} dx $$
