# Prove this limit doesn't exist (by contradiction)

I have to prove that $$\lim_{x \to \infty} \frac{3+x}{\sqrt{x}}$$ doesn't exist. But since I think it does exist I don't know what to do.

The clue was that we had to consider a proof by contradiction. How should I do this?

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Have you tried evaluating the expression for some values of $x$ such as 100, 10000, ... ? - and then asking if it could tend to a finite limit? –  Old John Nov 6 '12 at 21:01
First, what does $\displaystyle \lim_{x \to \infty} f(x) = L$ mean? Secondly, suppose that the limit equals some number $L$, and using the definition of the limit, derive a contradiction. –  JavaMan Nov 6 '12 at 21:01
John: Well since choosing larger values for x results in larger output I think the limit of this functions should be $$\infty$$ Java: L should become a fixed number? But then I can also state $$\lim_{x \to \infty} x$$ doesnt exist? –  Bob Nov 6 '12 at 21:05
There are some differences in standard usage. Some (books, courses, people) allow $\infty$ and $-\infty$ as possible limits. Some don't. If you have been asked to show that $\lim_{x\to\infty}\frac{3+x}{\sqrt{x}}$ doesn't exist, the person asking doesn't allow $\infty$ as a limit, at least in this course. So you need to show that there is no real number $b$ such that $\lim_{x\to\infty}\frac{3+x}{\sqrt{x}}=b$. –  André Nicolas Nov 6 '12 at 21:11
But isnt that obvious since $\infty$ is no real number? Ah well, I guess I then just can write something down with JavaMan's hint.... –  Bob Nov 6 '12 at 21:13

Suppose the limit exists and equals $L\in\mathbb R$. Then $\lim \sqrt{x}=\lim \frac{3+x}{\sqrt{x}}-\lim \frac{3}{\sqrt{x}}=L-0=L$, a contradiction since we know that $\sqrt{x}\to +\infty$.