anonymous one year ago can someone tell me how to prove for infinite limit with epsilon and delta or N and M?

1. abb0t

Yes, if you gave us an example, we could.

2. abb0t

What's the function?

3. anonymous

l$\lim_{x \rightarrow \infty}3x^{2}-x-2/5x^{2}+4x+1$

4. ganeshie8

$\lim_{x \rightarrow \infty}~\dfrac{3x^{2}-x-2}{5x^{2}+4x+1}$ like that ?

5. anonymous

Yes, like that

6. ganeshie8

so what does it evaluate to ?

7. ganeshie8

you need to know the limit before starting to prove it using epsilon-delta

8. anonymous

i know the limit it's 3/5

9. anonymous

but the prove is quite confusing.

10. ganeshie8

its easy once you know how to play the game

11. anonymous

i guess... so :)

12. ganeshie8

Let $$f(x) =\dfrac{3x^{2}-x-2}{5x^{2}+4x+1}$$ given any positive number $$\epsilon$$, you have to show that there exists some $$x$$ value, after which the difference between $$f(x)$$ and $$\dfrac{3}{5}$$ is less than $$\epsilon$$.

13. ganeshie8

Given any positive number $$\epsilon$$, prove that there exists some $$x$$ value beyond which below is true : $$\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\lt \epsilon$$

14. ganeshie8

simplify and try expressing $$x$$ in terms of $$\epsilon$$

15. ganeshie8

@kal1921

16. anonymous

simplify? sorry where is delta?

17. zzr0ck3r

Simplify so that you can find the bound on delta. You have to work backwards

18. ganeshie8

yeah what we're doing is just scratch work to express $$x$$ in terms of $$\epsilon$$

19. anonymous

$-\epsilon< equation<\epsilon$ i get this...what is next?

20. ganeshie8

for $$x\gt 1$$, we have : $$\left|\dfrac{3x^2-x-2}{5x^2+4x+1}-\dfrac{3}{5}\right|\\~\\ =\left|\dfrac{17x+13}{5(5x^2+4x+1)}\right|\\~\\ \lt\left|\dfrac{17x+13}{25x^2}\right|\\~\\ \lt\left|\dfrac{17x+17x}{25x^2}\right|\\~\\ =\left|\dfrac{34}{25x}\right|\\~\\$$ $$\dfrac{34}{25x} \lt \epsilon \implies x \gt \dfrac{34}{25\epsilon}$$ therefore $$x = \dfrac{34}{25\epsilon}$$ works, now you can start the proof.

21. anonymous

sorry, how can i relate x and delta... if it x wasn't approaching to infinity i would say /x-a/>delta

22. ganeshie8

The definition for limits at infinity is slightly different : $$\lim\limits_{x\to\infty}f(x)=L$$ means that for "every" $$\epsilon\gt 0$$, there exists some $$x$$ value, $$N$$, such that $$|f(x)-L|\lt\epsilon$$ for all $$x\gt N$$.

23. ganeshie8

In present problem, we have cooked up that $$N$$ to be $$\dfrac{34}{25\epsilon}$$

24. ganeshie8

you choose "any" positive $$\epsilon$$, the value $$|f(x)-\frac{3}{5}|$$ will always be less than that chosen $$\epsilon$$ whenever $$x$$ is greater than $$\dfrac{34}{25\epsilon}$$

25. ganeshie8

In other words, we can make $$f(x)$$ as close to $$\frac{3}{5}$$ as we want by making $$x$$ sufficiently large.

26. ganeshie8

I think this is same as the N and M definition that you were refering to in main question