Why does $\lim\limits_{x\to\infty} e^{-x^2} \int_x^{x+y}e^{t^2}dt$ differ if $y=\frac{1}{x}$, $y=\frac{\log{x}}{x}$, and $y=\frac{\log{x}}{2x}$? In a problem in Spivak's Calculus we discover the following
$$\lim\limits_{x\to\infty} e^{-x^2} \int_x^{x+\frac{1}{x}}e^{t^2}dt=0$$
$$\lim\limits_{x\to\infty} e^{-x^2} \int_x^{x+\frac{\log{x}}{x}}e^{t^2}dt=\infty$$
$$\lim\limits_{x\to\infty} e^{-x^2} \int_x^{x+\frac{\log{x}}{2x}}e^{t^2}dt=\frac{1}{2}$$
It seems like these limits are all measuring the same thing (but clearly they are not). The length of the integration interval goes to zero in all three cases.
I would have expected all three limits to be 0. Is there an intuitive (perhaps graphical) explanation of why the limits differ?
 A: This is a heuristic answer to understand the intuition of why such a thing is possible. Shift the intervals to $0$ by the translation $t\leftrightarrow t+x$
$$e^{-x^2}\int_x^{x+y}e^{t^2}\:dt = e^{-x^2}\int_0^ye^{(t+x)^2}\:dt = \int_0^ye^{t^2}e^{2tx}\:dt$$
The integrand blows up at the speed $e^{2yx}$ at the most - in the three cases given in the problem this gives us the speeds
$$\begin{cases} \text{Case I:} & e^{2} \\ \text{Case II:} & x^2 \\ \text{Case III:} & x \end{cases}$$
In the first case, the integrand at best doesn't grow at all, it's just a constant height times a function that approaches $1$. The shrinking outpaces the growth, so the limit is $0$.
In the other two cases, we have a quadratic speed and a linear speed. In this case, a quadratic speed was enough to outpace the shrinkage of the interval, but the linear growth happened to be "just right." As to why the linear growth happened to be just the right speed, the details of the limit calculation in Spivak say why. Intuitively if you can figure out not only a limit, but a rate of convergence to that limit, you can create interesting behaviors by making these things cancel out.
A: Let $y(x)=a\frac{\log(x)}{x}$. The limit is clearly $0$ when $a=0$. If $a>0$ then the limit is a ratio with the denominator that tends to $+\infty$. So we may apply L'Hopital's rule (and the Fundamental theorem of calculus),
$$\begin{align}
\lim_{x\to\infty} \frac{\int_x^{x+a\frac{\log(x)}{x}}e^{t^2}dt}{e^{x^2}} 
&=\lim_{x\to\infty}\frac{e^{(x+a\frac{\log(x)}{x})^2}(1+a\frac{1-\log(x)}{x^2})-e^{x^2}}{2xe^{x^2}}
\\
&=\lim_{x\to\infty}\frac{e^{2a\log(x)}(1+o(1))-1}{2x}\\
&=\lim_{x\to\infty}\frac{x^{2a}-1}{2x}=\begin{cases}
0 &\text{if $a<1/2$,}\\
1/2 &\text{if $a=1/2$,}\\
+\infty &\text{if $a>1/2$.}\\
\end{cases}
\end{align}$$
In a similar way you can show that for any $a\geq 0$,
$$\lim_{x\to\infty} e^{-x^2} \int_x^{x+\frac{a}{x}}e^{t^2}dt =0.$$
