ksaimouli
  • ksaimouli
limit
Mathematics
schrodinger
  • schrodinger
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ksaimouli
  • ksaimouli
\[\lim_{x \rightarrow {-\infty}}\frac{ 2x-1 }{ 1+2x }\]
ksaimouli
  • ksaimouli
when i divide by x i get |dw:1361724743582:dw|
ksaimouli
  • ksaimouli
and this leads me to 1 but the answer was -1

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ZeHanz
  • ZeHanz
OK, now what is lim x to inf of 1/x?
ksaimouli
  • ksaimouli
plz help with this
ksaimouli
  • ksaimouli
0
ksaimouli
  • ksaimouli
|dw:1361724809521:dw|
ksaimouli
  • ksaimouli
=1
ZeHanz
  • ZeHanz
So you get (2-0)/(0+2)=2/2=1
ksaimouli
  • ksaimouli
as this goes to - infin it is -1
ksaimouli
  • ksaimouli
i mean i got -1 bu the book says it is 1
ZeHanz
  • ZeHanz
No, it is not. It is always 1
ksaimouli
  • ksaimouli
why
ksaimouli
  • ksaimouli
as it goes to - infinity to should go to -1 right
ZeHanz
  • ZeHanz
Because you end up with (2x)/(2x) = 1
ksaimouli
  • ksaimouli
okay i got thx so no matter what after dividing we get the answer and i should not check the limit right
ZeHanz
  • ZeHanz
Just try a very big neg number, say x = -100000. Then the quotiens is: -200001/-199999, which is about 1
ZeHanz
  • ZeHanz
So you can even check the limit.
phi
  • phi
I am guessing a typo in the question or the answer.
ZeHanz
  • ZeHanz
@phi: me too
ZeHanz
  • ZeHanz
Remember: sometimes, even books are wrong ;)
phi
  • phi
especially math books. Proof readers miss all kinds of mistakes.
ksaimouli
  • ksaimouli
the answer is 1
ZeHanz
  • ZeHanz
Yes, @ksaimouli , you were right all the time!
ksaimouli
  • ksaimouli
thx
ZeHanz
  • ZeHanz
YW!

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