anonymous
  • anonymous
limit
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[\lim_{h \rightarrow 0}\frac{ \frac{ 4 }{ 2+2h }-2 }{ h }\]
anonymous
  • anonymous
|dw:1350448824402:dw|
anonymous
  • anonymous
limit approaches -2

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anonymous
  • anonymous
no it approaches 0
terenzreignz
  • terenzreignz
maybe reducing it all to just one fraction might help
anonymous
  • anonymous
|dw:1350449555332:dw| i know im doing something wrong
anonymous
  • anonymous
find a CD in the numerator .
anonymous
  • anonymous
\[\huge \frac{ \frac{ 4-2(2+2h) }{ 2+2h } }{h }\]
terenzreignz
  • terenzreignz
\[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+2h}-2}{h}\] is the same as \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+2h}-\frac{2(2+2h)}{2+2h}}{h}\] and further... \[\huge \lim_{h \rightarrow 0}-\frac{\frac{4h}{2+2h}}{h}\]
terenzreignz
  • terenzreignz
Now reduce it to one fraction, and do any cancellations, if possible
anonymous
  • anonymous
why is it \[\frac{ 4x }{ 2+2h }\]
anonymous
  • anonymous
its supposed to be (2+h) *
terenzreignz
  • terenzreignz
Looked at your first post... Ok, then \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-2}{h}\]\[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-\frac{2(2+h)}{2+h}}{h}\] \[\huge \lim_{h \rightarrow 0}\frac{\frac{4}{2+h}-\frac{4+2h}{2+h}}{h}\]
anonymous
  • anonymous
yup limit approaches to -2 as x approaches to zero
terenzreignz
  • terenzreignz
@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.
anonymous
  • anonymous
sorry typing error :$
anonymous
  • anonymous
|dw:1350450481516:dw| why do i get this :S this doesnt make sense because that would mean the answer is zero
anonymous
  • anonymous
like what am i doing wrong?
terenzreignz
  • terenzreignz
You were to multiply \[\frac{2+h}{2+h}\] to the -2 in the numerator; not to the entire fraction.
terenzreignz
  • terenzreignz
that among other things... But yeah, multiply the \[\frac{2+h}{2+h}\] only to the -2, and redo your work...
anonymous
  • anonymous
but that doesnt get rid of the fraction
anonymous
  • anonymous
hmm @terenzreignz, i didnt see the first posting of 2+h, i only saw 2+2h,, ththnx for reminding though...:D
terenzreignz
  • terenzreignz
We're not to get rid of it, only to make it simpler, ie, only one fraction bar.
anonymous
  • anonymous
|dw:1350450882525:dw| like that ?
terenzreignz
  • terenzreignz
That's right. Please continue.... :)
anonymous
  • anonymous
cross multiply?
anonymous
  • anonymous
nvm doesnt matter
terenzreignz
  • terenzreignz
Just distribute the 2.
anonymous
  • anonymous
|dw:1350451064789:dw|
terenzreignz
  • terenzreignz
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...
terenzreignz
  • terenzreignz
Not to mention you may have ignored and/or forgotten that there is another h in the denominator. Careful...
anonymous
  • anonymous
ohh alright !

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