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I've been given functions $f(x)=e^{x^2}$ and $F(x) = \int_{0}^{x} f(t) \,dt$. I have to show that $$\frac{f(x)}{x} \rightarrow \infty \quad \text{and} \quad F(x) \rightarrow \infty \quad \text{for} \quad x \rightarrow \infty$$ So far, all I have is that i first have to show that $f(x) / x$ converges to infinity and then use this to show that $F(x)$ does the same. How do I go about this?

My first thought was that $f(x)$ is an increasing function and therefore I could show that for some $M > 0$ we could find $x_0 > 0$ and show that all $f(x) / x > M$ for all $x > x_0$. So we would end up with something like this $$ \begin{equation*} \begin{split} f(x) / x & > M \\ e^{x^2} / x & > M \\ e^{x^2} & > Mx \\ x^2 & > \ln (Mx) \\ x^2 & > \ln (M) + \ln (x) \end{split} \end{equation*} $$ Am I completely on the wrong track here?

Fjeeds Arcade
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  • Partially wrong: the fact that $f$ is increasing does not help, and the sign $\iff$ is missing between each line of your list of inequalities. This sign is crucial. Better write only the direction which will be useful, i.e. $\Leftarrow$ (not $\Rightarrow$!). I.e. the last line is what remains to prove in order to ensure the first line. More precisely, you must prove that $\forall M>0\quad\exists x_0>0\quad x^2>\ln M+\ln x.$ – Anne Bauval Mar 15 '23 at 15:15
  • "Converges to infinity" is a term not used in mathematics. A function may converge to some number, say $C$. But if the functions keeps increasing and goes to infinity, mathematicians say: "The function diverges". Diverge is the opposite of converge. – M. Wind Mar 15 '23 at 16:04
  • Sorry, you are correct. I should have written it as $e^{x^2} \rightarrow \infty$ instead. – Fjeeds Arcade Mar 15 '23 at 16:05

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The easiest way to show both statements is to estimate $f(x) = e^{x^2}$ from below by some easier function that also tends to infinity stronger than $x$.

For example, $e^{x^2} \geq x^2$ for all $x\in\mathbb{R}$ (this follows from the basic inequality $e^x>x$ for all $x>0$, which can e.g. be proven by using the Taylor series of $e^x$ for $x>0$), so $$\lim\limits_{x\rightarrow\infty}\frac{e^{x^2}}{x} \geq \lim\limits_{x\rightarrow\infty}\frac{x^2}{x} = \lim\limits_{x\rightarrow\infty} x = \infty$$

and $$F(x)=\int\limits_0^x f(t)dt \geq \int\limits_0^x t^2dt = \frac{1}{3}x^3,$$ which obviously satisfies $\lim\limits_{x\rightarrow\infty} F(x) = \infty$.

Nuke_Gunray
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    Remark: A stronger inequality than the one used is $e^x \geq x + 1$, which might be more useful in other problems. Besides, why not just apply L'Hopital's rule for the first limit? – Morărescu Mihnea Mar 15 '23 at 15:51
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    @MorărescuMihnea You are right, that formula is more potent. Considering your second remark: Of course L'Hospital is very usefull in this case, but I think for learners of calculus, it's important to learn how to do estimates from an early stage on, since elsewhise they are kind of lost when faced with problems where L'Hospital is not applicable, or just try to apply it every time they encounter a fractional limit, no matter if it is applicable or not. – Nuke_Gunray Mar 15 '23 at 15:54
  • This makes perfect sense to me! Would this also be useful when I later have to find the limit for $\frac{F(x)}{f(x)/x}$? – Fjeeds Arcade Mar 15 '23 at 16:35
  • @FjeedsArcade Only one of them: If you have a fraction $\frac{a}{b}$, and know that $a\geq \tilde a$ and $b\geq \tilde b$, all you can say is that either $\frac{a}{b}\geq \frac{\tilde a}{b}$ or $\frac{a}{b}\leq \frac{a}{\tilde b}$. To estimate both numerator and denominator at the same time, you either need two inequalites like $a\geq \tilde a$, $b\leq \tilde b$ or $a\leq \tilde a$, $b\geq \tilde b$. But I think a different approach would be faster - for example L'Hospital. – Nuke_Gunray Mar 15 '23 at 17:40