If limit of a function exists does it imply that function is bounded?

Question

If for a real function $f$, the limit $\lim_{x\to \infty}f(x)$ exists, then is it true that $f$ is bounded ?

Note that it is limit of a function, not sequence.

I don't think it is true but found that in some book.

Actually, the claim in the book is that as $\lim_{t\to \infty}\int_{0}^{t}\frac{\sin x}{x}dx = \frac{\pi}{2}$, $\sup_{t} |\int_{0}^{t}\frac{\sin x}{x}dx| < \infty$. I don't know why is this true ?

Answer 1

It's not true. Think of the function \begin{align} f : \mathbb R \to \mathbb R : x \mapsto \begin{cases}0 & x = 0\\ \frac{1}{x}& \text{else} \end{cases} \end{align}

You might mean "is it 'bounded at \infty'," which might mean something like this: "as you approach infinity, there's some number $M$ such that $f(x) \le M$." In that case, the answer is "yes": you can just pick $M = |L| + 1$, where $L$ is the limit at infinity.

Finally, if you assume that $f$ is continuous, then the claim actually is true, by a combination of the last paragrapth and the fact that for $x \le M$, you're looking at a continuous function on a closed interval, which achieves both its max and its min.

Answer 2

You can think of the function $f(x)=\frac{1}{x}$: though $\lim_{x\rightarrow \infty} \frac{1}{x}=0$ it isn't bounded because $\lim_{x\rightarrow 0} \frac{1}{x}=\infty$.

Answer 3

Not true for e.g. $f(x)=e^{-x}$ which is exists as $x→∞$ but what about $f(x)$ when $x→-∞$?