3
$\begingroup$

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

Thanks in advance!

$\endgroup$
  • 3
    $\begingroup$ The second is not very different from the first. Have you written down what $f(x) - L$ is in that situation? $\endgroup$ – Dylan Moreland Sep 12 '11 at 5:53
4
$\begingroup$

This may help you get started:

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 $\varepsilon=1$)

$\endgroup$
1
$\begingroup$

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, $N=\left\lceil \frac{10}{\varepsilon}\right\rceil$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.