suneja Group Title lim x^2+x-11=9 x tends to 4 prove tis by formal definition of limit one year ago one year ago

1. Jeffrey_Calderon Group Title

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

2. suneja Group Title

sure i can ... wanted to jst chk

3. heedcom Group Title

just plug 4 into the equation

4. nickhouraney Group Title

@suneja using epsilon and delta right

5. suneja Group Title

@nickhouraney ya can u show dat here

6. rerukumash Group Title

wats lim anyone pls help

7. suneja Group Title

lim is limit

8. nickhouraney Group Title

ya i can show u

9. rerukumash Group Title

ok where is it used

10. rerukumash Group Title

11. nickhouraney Group Title

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

12. satellite73 Group Title

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

13. suneja Group Title

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 Group Title

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 Group Title

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 Group Title

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 Group Title

@nickhouraney thanks

18. nickhouraney Group Title

well lets see your proof

19. suneja Group Title

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

20. satellite73 Group Title

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 Group Title

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 Group Title

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

23. satellite73 Group Title

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 Group Title

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 Group Title

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 Group Title

not square root, absolute value is what i meant

27. satellite73 Group Title

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 Group Title

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

29. satellite73 Group Title

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 Group Title

got it ..thanks all

31. suneja Group Title

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