## suneja 3 years ago lim x^2+x-11=9 x tends to 4 prove tis by formal definition of limit

1. Jeffrey_Calderon

that's quite long to type. haha. you can do it dude! :)

2. suneja

sure i can ... wanted to jst chk

3. heedcom

just plug 4 into the equation

4. nickhouraney

@suneja using epsilon and delta right

5. suneja

@nickhouraney ya can u show dat here

6. rerukumash

wats lim anyone pls help

7. suneja

lim is limit

8. nickhouraney

ya i can show u

9. rerukumash

ok where is it used

10. rerukumash

11. nickhouraney

|x-a| |x-4| |f(x)-L|<epsilon | (x^2+x-11) - 9| start from here

12. satellite73

start with $|x^2-x-20|<\epsilon$ and see what you need for $$|x-4|$$ to make it happen

13. suneja

mod (x^2-x-11)-9 < epsilon wenever 0<mod x-4 < delta tis is how v start then go by simplifying left inequality to find a suitable value for delta then verify that choice of delta i hav a prob in verification .. need help der

14. nickhouraney

we basically need to take this | (x^2+x-11) - 9| < epsilon and make it |x-4|<epsilon start by simplifying | (x^2+x-11) - 9| that will turn to | x^2+x-11 - 9| < epsilon

15. suneja

ya then u hav to prove that "choice " of delta wic u get by solving the 1st part like u said can u show me 2nd part of tis prob

16. nickhouraney

for the proof you would just go in reverse basically. if |x-4| < delta then |x-4| < (your choice for epsilon) then show how this will lead to |f(x)-L| < epsilon

17. suneja

@nickhouraney thanks

18. nickhouraney

19. suneja

i've done exactly like but was lil confused in 2nd half of it! neways thanks

20. satellite73

oh yes it is. okay you should start with $|x^2+x-20|<\epsilon$ then factor to get $|(x-4)(x+5)|<\epsilon$ you have control over $$|x-4|$$ so you just need a bound for $$|x+5|$$

21. satellite73

trick is to say that since you are taking the limit as $$x\to 4$$ you can assert that say $$3<x<5$$ so the largest $$|x+5|$$ can be is 10

22. suneja

wat do u mean by " a bound for $\left| x+5 \right|$

23. satellite73

ok lets go slow you get to say how large $$|x-4|$$ can be, that is you get to pick your $$\delta$$ so that $$|x-4|$$ is smaller than $$\delta$$

24. satellite73

now you are looking at the product inside the square root, you have $$|(x-4)(x+5)|$$ and you want to make this smaller than $$\epsilon$$

25. satellite73

but $$|(x-4)(x+5)|=|x-4||x+5|$$ you can make the first term as small as you like, but you cannot make $$|x+5|$$ small

26. satellite73

not square root, absolute value is what i meant

27. satellite73

so the question is, how big can $$|x+5|$$ be? well it can be really really large, but don't forget you are making $$x$$ close to 4

28. satellite73

so you can assert that since $$x$$ is close to 4, it is certainly less that say 5, making $$|x+5|<10$$

29. satellite73

now you want $|(x-4)(x+5)|<\epsilon$ if you make $$\delta=\frac{\epsilon}{10}$$ and $$3<x<5$$ then you know $|(x-4)(x+5)|<10|x-4|<10\times \frac{\epsilon}{10}=\epsilon$

30. suneja

got it ..thanks all

31. suneja

$-\infty \rightarrow +\infty$ is not a bound right?