Evaluate the limit of $\lim_{x \to \infty} \left ( \frac{1}{x} \int_{0}^{x} e^{t^{2}} dt \right )$ Evaluate the limit of $$ \lim_{x \to \infty} \left ( \frac{1}{x} \int_{0}^{x} e^{t^{2}} dt \right ) $$
Solution:
based on Leibniz's Rule,
$$ \require{cancel} \cancel{\int_{0}^{x}e^{t^{2}}dt = e^{x^{2}} - e^{0^{2}} + \int_{0}^{x} \left(\frac{\mathrm{d}}{\mathrm{d}x} e^{t^{2}}\right) dt} $$
$$ \int_{0}^{x}e^{t^{2}}dt = e^{x^{2}} - e^{0^{2}} + \int_{0}^{x} \left(\frac{\mathrm{∂}}{\mathrm{∂}x} e^{t^{2}}\right) dt $$
and, since it is dx, can we treat t as a constant? So, $$ \frac{\mathrm{d}}{\mathrm{d}x} e^{t^{2}} = 0 $$
therefore we have,
$$ \frac{e^{x^{2}} - 1}{x}  $$
by apply L'Hôpital's rule,
$$ \frac{e^{x^{2}}2x - 0}{1} $$
so, when $x\rightarrow \infty$, the whole expression tends to infinity as well.
 A: Note that $f(t)=e^{t^2}$ is a positive function which grows 'very quickly' so that the area under the curve, $\int_0^x e^{t^2} \;dt$ should increase very quickly - much faster than that for a line. So we suspect the limit should be infinite. Let's try to make sense of this idea.
We know the first few terms of the Taylor Series for $f(t)= e^{t^2}$, namely $1+t^2 + O(t^4)$. [If you do not know this yet, do not worry.] So let's show that $e^{t^2}$ is always bigger than $g(t)= 1+t^2$ for $t \geq 1$. We know that $f(1)= e \approx 2.71828$ and $g(1)= 2$ so that $f(1)>g(1)$. Now $f'(t)= 2t e^{t^2}$ and $g'(t)= 2t$. But because $e>1$, we know that $f'(t)= 2te^{t^2}>2 t \cdot 1= 2t= g'(t)$. Therefore, $f(t) > g(t)$ for all $t \geq 1$.
Now because $f(t) \geq 0$, we know $\int_0^1 e^{t^2} \;dt \geq 0$. But then for $x>1$,
$$
\int_0^x e^{t^2} \;dt= \int_0^1 e^{t^2} \;dt + \int_1^x e^{t^2} \;dt \geq \int_1^x e^{t^2} \; dt
$$
Now we know that
$$
\int_1^x g(t) \;dt= \int_1^x (1+t^2) \;dt= \dfrac{x^3 + 3x-4}{3}
$$
But then we have
$$
\int_0^x e^{t^2} \;dt \geq \int_1^x (1+t^2) \;dt= \dfrac{x^3 + 3x-4}{3}
$$
But
$$
\lim_{x \to \infty} \dfrac{\dfrac{x^3 + 3x-4}{3}}{x}= \infty
$$
This shows that
$$
\lim_{x \to \infty} \dfrac{1}{x} \int_0^x e^{t^2} \;dt= \infty
$$
Note: You can use l'Hopital's rule as well. My opting for the solution above is that with 'non-obvious' l'Hopital problems, students often forget to check the underlying assumptions for l'Hopitals and then either have an incomplete solution or an incorrect one. Open your textbook and check the assumptions for l'Hopital's for the functions $f(x)= \int_0^x e^{t^2} \;dt$ and $g(x)= x$. Then from the comments above, we know that $\lim_{x \to \infty} f(x)= \infty$ and it is obvious that $\lim_{x \to \infty} g(x)= \infty$. Then using l'Hopital's rule (and the Fundamental Theorem of Calculus for $f'(x)$), we have
$$
\lim_{x \to \infty} \dfrac{\int_0^x e^{t^2} \;dt}{x} \stackrel{\text{L.H.}}{=} \lim_{x \to \infty} \dfrac{\frac{d}{dx} \int_0^x e^{t^2} \;dt}{\frac{d}{dx} x}= \lim_{x \to \infty} \dfrac{e^{x^2}}{1}= \infty
$$
