Is the limit of a strictly increasing function always $\infty$?

If $$f(x)$$ is strictly increasing, is $$\lim_{x\to\infty} f(x) = \infty$$ ? Also is $$\lim_{x\to -\infty} = 0?$$ I think the answer is yes. A good example is $$e^{x}$$. I don't know how to show this claim though. Is it true?

• No, if the function is bounded the limit is finite. Consider $f(x) =-1/x$ as an example. – Paramanand Singh Nov 12 '18 at 6:22
• Then how can I show $\lim_{x\to\infty} e^{x} = \infty$ ? – Dillain Smith Nov 12 '18 at 6:25
• @DillainSmith You can prove it directly from the definition of $\lim_{x\to\infty}f(x)=\infty$. – Arthur Nov 12 '18 at 6:26

Use the following theorem:

Theorem: Let $$f:[a, \infty) \to\mathbb{R}$$ be a function. If $$f$$ is increasing on $$[a, \infty)$$ then either $$\lim_{x\to\infty} f(x)$$ exists or $$f(x) \to\infty$$ as $$x\to\infty$$.

The proof can be given by analyzing the range $$A=f([a, \infty))$$ of $$f$$. If $$A$$ is bounded above (meaning every member of $$A$$ is less than some fixed number) then $$L=\sup A$$ exists and one can prove that $$f(x) \to L$$ as $$x\to\infty$$ (prove this!).

On the other hand if $$A$$ is unbounded above ie for any given $$K>0$$ there is a value $$x_0\geq a$$ such that $$f(x_0)>K$$ then by increasing nature of $$f$$ we have $$f(x) \geq f(x_0)>K$$ for all $$x>x_0$$. Thus $$f(x) \to \infty$$ as $$x\to\infty$$.

For $$f(x) =e^x$$ the situation is very simple as it satisfies the inequality $$e^x\geq x+1>x$$ for all $$x$$. Hence given any $$K>0$$ there is an $$x_0$$ namely $$x_0=K$$ such that $$f(x)=e^x>x>x_0=K$$ for all $$x>x_0$$. And thus $$f(x) \to\infty$$ as $$x\to\infty$$.

Hint: What about the $$\tanh$$ function?

You know what is $$\lim\limits_{x\to-\infty}e^x?$$What about $$\lim\limits_{x\to\infty}-e^{-x}?$$ And what about $$\lim\limits_{x\to-\infty}-e^{-x}?$$

What about for $$x>0$$

$$f(x)=1-\frac1x\implies f’(x)=\frac1{x^2}>0$$