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anonymous

  • one year ago

Can someone please explain the solution of the following limit to me? I'd really appreciate it!

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  1. anonymous
    • one year ago
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  2. AravindG
    • one year ago
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    Well first to check if the limit exist we plug in -1

  3. AravindG
    • one year ago
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    Here we get 0/0 form so we try to factorize the denominator and see if any cancellation is possible with numerator. In this case while factoring we got an (X+1) in denominator that we could cancel out in numerator.

  4. anonymous
    • one year ago
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    Right, I understand everything until it brings out a 3/2 from the limit

  5. AravindG
    • one year ago
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    You can see that once cancellation is done after factorizing the limits tends to a value.

  6. AravindG
    • one year ago
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    3/2? I am getting 1/4

  7. AravindG
    • one year ago
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    \[\dfrac{1}{(-1)(2)(-2)}\]

  8. anonymous
    • one year ago
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    Yeah i mean here... \[\frac{ 3 }{ 2 } \lim _{x \rightarrow1}\frac{ x-1 }{ \left| x-1 \right| }\]

  9. anonymous
    • one year ago
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    where does the 3/2 come from?

  10. AravindG
    • one year ago
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    Is this the same question?

  11. anonymous
    • one year ago
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    yeah, check the solution in the attachment

  12. anonymous
    • one year ago
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    Ohh my bad wrong attachment haha wait a sec

  13. AravindG
    • one year ago
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    I dont see 3/2

  14. AravindG
    • one year ago
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    :|

  15. anonymous
    • one year ago
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  16. anonymous
    • one year ago
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    in this one XD sorry lol

  17. AravindG
    • one year ago
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    (x+2)/(x+1)=(1+2)/(1+1)=3/2

  18. anonymous
    • one year ago
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    so why isn't x plugged in for the rest of the equation?

  19. AravindG
    • one year ago
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    They split the limit as lim xto1 (x+2)/(x+1) times lim xto1 x-1/|x-1|

  20. AravindG
    • one year ago
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    lim(g(x)f(x))=lim(g(x)) times lim(f(x)) Thats what they used

  21. AravindG
    • one year ago
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    Get it?

  22. anonymous
    • one year ago
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    Ohhh yeah now i get it

  23. AravindG
    • one year ago
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    well done :)

  24. anonymous
    • one year ago
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    but why? since there's an absolute value?

  25. AravindG
    • one year ago
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    You see g(x)=lim xto1 (x+2)/(x+1) is a limit that exists but for f(x) we needed more work so we took out known limit outside and dealt with f(x).

  26. AravindG
    • one year ago
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    Yeah absolute value means we need to observe the function under regions.

  27. anonymous
    • one year ago
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    can't i just cancel out the two (x-1) ?

  28. AravindG
    • one year ago
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    Nope.

  29. anonymous
    • one year ago
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    and just get 3/2

  30. AravindG
    • one year ago
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    |x-1|=x-1 when (x-1) >0 |x-1|=-(x-1) when (x-1)<0 This comes from definition of absolute value function. |x|=x when x>0 |x|=-x when x<0

  31. AravindG
    • one year ago
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    So the limit f(x) shows two values in 2 regions.

  32. anonymous
    • one year ago
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    So what if we find 3/2 and say that since its an absolute value function, its either 3/2 or -3/2 so the limit does not exist

  33. AravindG
    • one year ago
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    Limit exists for absolute value function. The only magic is that it exhibits different limits in different regions.

  34. AravindG
    • one year ago
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    That is why when we have absolute value function we give the answer in regions.

  35. anonymous
    • one year ago
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    Mhmm i understand...

  36. anonymous
    • one year ago
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    Im just thinking that I wouldn't have initially thought of approaching the problem like that (splitting it to 2 limits)

  37. AravindG
    • one year ago
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    Notice that you could do the problem without splitting also. We do the same operation on f(x) with the g(x) part remaining a person with no specific role. So to make it more appealing we split the limit.

  38. AravindG
    • one year ago
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    Properties like these are used by those who approach the problem rationally. It makes it more elegant.

  39. anonymous
    • one year ago
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    Oh okay, that makes sense

  40. anonymous
    • one year ago
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    you see at the end it breaks the lim into one with x -> 1+ and x->1- ? Would it be correct if i do that from the very start and compute those 2 limits separately?

  41. anonymous
    • one year ago
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    That would eventually give me the same answer right?

  42. AravindG
    • one year ago
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    Of course it would be right. The limit should be answered in intervals and it is in the hands of the student to decide when to split it to intervals either in the end or during the start.

  43. AravindG
    • one year ago
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    You could do all these cases separately and evaluate it in your notebook to make it 100% clear.

  44. AravindG
    • one year ago
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    That way you get more than one way to approach a problem.

  45. anonymous
    • one year ago
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    Awesome! :D Thanks so much @AravindG !!! I bugged you a lot haha I'm sorry! :D

  46. AravindG
    • one year ago
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    haha I like being bugged. It means the learner is a good student :)

  47. anonymous
    • one year ago
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    Lol Thanks!! :D i appreciate it!

  48. AravindG
    • one year ago
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    yw ;)

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