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burhan101

  • 3 years ago

limit

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  1. burhan101
    • 3 years ago
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    \[\lim_{h \rightarrow 0}\frac{ \frac{ 4 }{ 2+2h }-2 }{ h }\]

  2. burhan101
    • 3 years ago
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    |dw:1350448824402:dw|

  3. mark_o.
    • 3 years ago
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    limit approaches -2

  4. burhan101
    • 3 years ago
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    no it approaches 0

  5. terenzreignz
    • 3 years ago
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    maybe reducing it all to just one fraction might help

  6. burhan101
    • 3 years ago
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    |dw:1350449555332:dw| i know im doing something wrong

  7. Algebraic!
    • 3 years ago
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    find a CD in the numerator .

  8. Algebraic!
    • 3 years ago
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    \[\huge \frac{ \frac{ 4-2(2+2h) }{ 2+2h } }{h }\]

  9. terenzreignz
    • 3 years ago
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    \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+2h}-2}{h}\] is the same as \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+2h}-\frac{2(2+2h)}{2+2h}}{h}\] and further... \[\huge \lim_{h \rightarrow 0}-\frac{\frac{4h}{2+2h}}{h}\]

  10. terenzreignz
    • 3 years ago
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    Now reduce it to one fraction, and do any cancellations, if possible

  11. burhan101
    • 3 years ago
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    why is it \[\frac{ 4x }{ 2+2h }\]

  12. burhan101
    • 3 years ago
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    its supposed to be (2+h) *

  13. terenzreignz
    • 3 years ago
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    Looked at your first post... Ok, then \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-2}{h}\]\[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-\frac{2(2+h)}{2+h}}{h}\] \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-\frac{4+2h}{2+h}}{h}\]

  14. mark_o.
    • 3 years ago
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    yup limit approaches to -2 as x approaches to zero

  15. terenzreignz
    • 3 years ago
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    @mark_o. You might want to reconsider..., given that there was a correction; it's 2+h, not 2+2h as was in the first post.

  16. burhan101
    • 3 years ago
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    sorry typing error :$

  17. burhan101
    • 3 years ago
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    |dw:1350450481516:dw| why do i get this :S this doesnt make sense because that would mean the answer is zero

  18. burhan101
    • 3 years ago
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    like what am i doing wrong?

  19. terenzreignz
    • 3 years ago
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    You were to multiply \[\frac{2+h}{2+h}\] to the -2 in the numerator; not to the entire fraction.

  20. terenzreignz
    • 3 years ago
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    that among other things... But yeah, multiply the \[\frac{2+h}{2+h}\] only to the -2, and redo your work...

  21. burhan101
    • 3 years ago
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    but that doesnt get rid of the fraction

  22. mark_o.
    • 3 years ago
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    hmm @terenzreignz, i didnt see the first posting of 2+h, i only saw 2+2h,, ththnx for reminding though...:D

  23. terenzreignz
    • 3 years ago
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    We're not to get rid of it, only to make it simpler, ie, only one fraction bar.

  24. burhan101
    • 3 years ago
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    |dw:1350450882525:dw| like that ?

  25. terenzreignz
    • 3 years ago
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    That's right. Please continue.... :)

  26. burhan101
    • 3 years ago
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    cross multiply?

  27. burhan101
    • 3 years ago
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    nvm doesnt matter

  28. terenzreignz
    • 3 years ago
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    Just distribute the 2.

  29. burhan101
    • 3 years ago
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    |dw:1350451064789:dw|

  30. terenzreignz
    • 3 years ago
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    ok... I think I know where you're getting at... Be careful, the way you wrote it is full of... difficulties. On top, you should have 4 - (4 + 2h) Groupings can make a world of difference, so you better specify which operations are prioritised...

  31. terenzreignz
    • 3 years ago
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    Not to mention you may have ignored and/or forgotten that there is another h in the denominator. Careful...

  32. burhan101
    • 3 years ago
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    ohh alright !

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