# How to prove $\lim\limits_{x\to \infty}3x=\infty$?

How to prove $\lim\limits_{x\to \infty}3x=\infty$?

First I am not sure about formal definition of $\lim\limits_{x\to \infty}f(x)=\infty$, I guess $\forall K\in \Bbb{R},\exists N\in \Bbb{R}:x\gt N\implies f(x)\gt K$

If that's the case,

Let $K\in \Bbb{R}$, let $N=\frac{K}{3}$, then $x\gt \frac{K}{3}\implies 3x\gt K$.

I am sure it's not this simple. Could someone givea valid one?

• It is that simple, because $3x$ is very simple. – TSF Apr 18 '16 at 0:52
• Even just showing that $3(\infty)$ is probably convincing enough for any mathematician – q.Then Apr 18 '16 at 0:56
• $x \to 3x$ is a strictly increasing surjection. $3x$ can therefore be made arbitrarily large by making $x$ sufficiently large. – MathematicsStudent1122 Apr 18 '16 at 1:14
• @McCheng that is an injection, not a surjection – JMoravitz Jun 23 '16 at 0:10
• If it the limit converges, then it is bounded above. Therefore, if it is not bounded above, it must diverge to infinity; hence the definition you give for divergence, leading to your simple (and valid) result. – AloneAndConfused Feb 24 '17 at 13:40

Forget about the question for a moment...

Lets define $f(x)=3x$.

Now, start putting values into it.

$$f(1)=3$$

$$f(9)=27$$

$$f(99)=297$$

$$f(999)=2997$$

$$f(9999)=29997$$

$...$ and so on.

As you can see, the larger the value of $x$ gets, the larger is the value of $f(x)$. So, as $x$ approaches $\infty$, $f(x)=3x$ must also approach $\infty$.

Here is a graph for $f(x)$ with a "large scale" :

I think it must be pretty intuitive.

Otherwise, you can see that

$$\lim \limits_{x \rightarrow \infty} f(x) = \lim \limits_{x \rightarrow \infty} 3x = 3\Bigg(\lim \limits_{x \rightarrow \infty} x\Bigg) = 3 \times \infty = \infty .$$

Hope this helps :-)