# Limit proof as $x$ goes to infinity

Trying to help my girlfriend since we're both in Calculus this semester, but my class never went this in depth into limit proofs. A little help so I can pass it on? (:

The limit at infinity $\lim_{x\to\infty} f(x) = L$ means that for any $\varepsilon > 0$, there exists $N > 0$ such that

$|f(x) - L| < \varepsilon$ whenever $x > N$.

Use this definition to prove the following statements.

1. $\lim\limits_{x \to +\infty} \frac{10}{x} = 0.$
2. $\lim\limits_{x \to +\infty} \frac{2x+1}{x} = 2.$

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The second is not very different from the first. Have you written down what $f(x) - L$ is in that situation? – Dylan Moreland Sep 12 '11 at 5:53

Can you find a number $N$ (greater than 0) such that, for any number $x>N$, $$\left|\;\frac{10}{x}-0\;\right|=\left|\;\frac{10}{x}\;\right|=\frac{10}{x}<1\quad\quad?$$ (so here we are looking at $\varepsilon=1$)
For both questions you have to do the same thing. For example, in 1, you are given a positive $\varepsilon$ and you have to find $N$ such that $\left|\;\frac{10}{x}-0\;\right|=\left|\;\frac{10}{x}\;\right|=\frac{10}{x}<\varepsilon$ holds for all $x>N$. Since it clearly holds for all $x>\frac{10}{\varepsilon}$, you can take, for example, $N=\left\lceil \frac{10}{\varepsilon}\right\rceil$.