# If $f(x) \rightarrow c > 0$ when $x \rightarrow \infty$, is it true that its anti-derivative $F(x) \rightarrow \infty$ when $x \rightarrow \infty$?

If $$f$$ is a continous function and $$f(x) \rightarrow c > 0$$ when $$x \rightarrow \infty$$, is it true that the function's anti-derivative $$F(x) \rightarrow \infty$$ when $$x \rightarrow \infty$$?

Intuitively, I would say that this is true because I believe that there should be an infinitely large area constrained by the graph of $$f(x)$$ (which is above the positive $$x-axis$$) and the positive $$x$$-axis as $$x \rightarrow \infty$$. However, I cannot prove this. Is my intuition correct and might a more rigorous argument look like?

• A slightly stronger result holds under the conditions of the question. By L'Hospital's Rule $F(x) /x\to c$ as $x\to \infty$ and your desired conclusion follows. – Paramanand Singh Feb 18 at 5:23

Its true and a rigourous argument is just across the road. Pick $$M$$ such that for $$x>M$$, we have the bound $$f(x) > c/2>0.$$ It is enough to show that $$\lim_{y\to\infty} \int_M^y f = \infty$$. And this is true, because...
You know that there is some $$X \in \mathbb{R}$$ such that for all $$x > X$$, $$f(x) \in [\frac{1}{2}c, \frac{3}{2}c]$$, so $$f(x) > \frac{c}{2} > 0$$, since $$f(x)$$ comes arbitrarily close to $$c$$ if $$x$$ gets big enough. Hence for all $$x > X$$: $$\int_0^x f(s) ds = \int_0^X f(s) ds + \int_X^x f(s) ds > \int_0^X f(s) ds + (x-X)\frac{c}{2}.$$ As $$x \to \infty$$, the second term blows up, so $$F(x) \to \infty$$ as $$x\to \infty$$.
Because $$f$$ converges to $$c$$, there exists $$u\in\mathbb R$$ such that if $$x\geq u$$, then $$f(x)\geq \frac c 2$$. Then $$F(x)=F(u)+\int_u^x f(t)dt \geq F(u) +\int_u^x \frac{c}{2}dt=F(u) +\frac c 2 (x-u)$$ And the latter $$\rightarrow+\infty$$.
Yes, that is true. One way to see it is using the FTC: you can say that $$f(x)>c/2$$ for large enough $$x$$, say $$x>a$$) (by the definition of limit) and then $$F(x)=F(a)+\int\limits_a^xF'(x)\,dx=F(a)+\int\limits_a^x f(x)\,dx$$ $$\ge F(a)+\int\limits_a^x \frac{c}{2}\,dx \ge F(a)+\frac{c}{2}(x-a)$$ so $$F$$ grows at least linearly at infinity.