AmTran_Bus
  • AmTran_Bus
Limit question?
Mathematics
katieb
  • katieb
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AmTran_Bus
  • AmTran_Bus
|dw:1436286205012:dw|
AmTran_Bus
  • AmTran_Bus
Um. Here are my choices but Im getting DNE
AmTran_Bus
  • AmTran_Bus
Just had an idea

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AmTran_Bus
  • AmTran_Bus
What in the world someone help!http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+%28e%5Ex-6%29%2Fsin+4x
AmTran_Bus
  • AmTran_Bus
The problem does not say from the left or right!!!!!!!!! So my DNE is right but what do I do?
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{e^x-6}{\sin(4x)}}\) ?
AmTran_Bus
  • AmTran_Bus
Correct
welshfella
  • welshfella
would l'hopitals rule be helpful here?
AmTran_Bus
  • AmTran_Bus
Good thought. Let me try it
SolomonZelman
  • SolomonZelman
can't do it yet
SolomonZelman
  • SolomonZelman
it is not 0/0 when you plug in x=0
welshfella
  • welshfella
right - only the denoiminator = 0 l'hopitals no help
AmTran_Bus
  • AmTran_Bus
Oh yea, must be inf/inf or 0/0
welshfella
  • welshfella
as x approaches 0 sin 4x approaches 0 and e^x - 6 approaches -5 so can we then say that the limit is - infinity?
welshfella
  • welshfella
- approaching from above 0
SolomonZelman
  • SolomonZelman
you are right, the limit actually doesn't exst.
AmTran_Bus
  • AmTran_Bus
Yea, thats what has thrown me off for a long time. But wolfram treats it as 0+ and 0- and and the answer choices have both neg inf and pos inf
SolomonZelman
  • SolomonZelman
even one sides limits diverge
SolomonZelman
  • SolomonZelman
when you get a limit that is equal to +∞ or -∞, that means the limit does NOT exist. (It is infinite - not limited)
AmTran_Bus
  • AmTran_Bus
But it is an ans choice!
SolomonZelman
  • SolomonZelman
Well, for two sides limit, it DNE
AmTran_Bus
  • AmTran_Bus
I agree
SolomonZelman
  • SolomonZelman
from the right side it is ∞, from the left side it is -∞
AmTran_Bus
  • AmTran_Bus
dumb problem and ans choices
SolomonZelman
  • SolomonZelman
do you have an answer choice does not exist?
AmTran_Bus
  • AmTran_Bus
welshfella
  • welshfella
yes near the y axis the graph looks like |dw:1436287092439:dw|
AmTran_Bus
  • AmTran_Bus
AmTran_Bus
  • AmTran_Bus
@welshfella so do you agree DNE?
welshfella
  • welshfella
yes
AmTran_Bus
  • AmTran_Bus
Well thanks for all your help! I may email prof
AmTran_Bus
  • AmTran_Bus
Well, I asked on yahoo ans and they said -inf. I just dont see it but gotta mark something @hartnn @SolomonZelman @welshfella
hartnn
  • hartnn
limit shouldn't exist sin 4x is negative when x<0 and it is positive when x>0 hence Left hand limit would not = right hand limit
AmTran_Bus
  • AmTran_Bus
I agree hartnn. Just these ans choices giving fit.
SolomonZelman
  • SolomonZelman
ok, maybe the question is a little bit different?
AmTran_Bus
  • AmTran_Bus
??
SolomonZelman
  • SolomonZelman
Are you sure it is exactly saying \(\Large\color{slate}{\displaystyle\lim_{x \rightarrow ~0}\frac{e^x-6}{\sin(4x)}}\)
SolomonZelman
  • SolomonZelman
well, then this limit doesn't exist.
SolomonZelman
  • SolomonZelman
I mean if limit=∞ or -∞ it doesn't exist (either way)
hartnn
  • hartnn
people saying -infinity are assuming that sin4x = 0 e^x -6 will be negative
SolomonZelman
  • SolomonZelman
ohh...
SolomonZelman
  • SolomonZelman
yeah, should have noticed that:)
SolomonZelman
  • SolomonZelman
Bus?
AmTran_Bus
  • AmTran_Bus
I'm here! So is it -inf after all?
SolomonZelman
  • SolomonZelman
yes
AmTran_Bus
  • AmTran_Bus
Sin 4x=0?
SolomonZelman
  • SolomonZelman
yes
hartnn
  • hartnn
no lol, the limit is still DNE. In yahoo answer, they might have said -inf as they might have assumed sin4x =0 ... that was my point, but sin 4x is not actually 0
SolomonZelman
  • SolomonZelman
well, if you take the limit of sin(4x) as x→0 alone, then it is 0
AmTran_Bus
  • AmTran_Bus
Story of my life...
hartnn
  • hartnn
why would you do that lol...
SolomonZelman
  • SolomonZelman
1 D, XD = (1+x)D
SolomonZelman
  • SolomonZelman
jk
SolomonZelman
  • SolomonZelman
idk, hartnn
AmTran_Bus
  • AmTran_Bus
Gonna email prof
SolomonZelman
  • SolomonZelman
the limit is indeterminate, beause you are trying to divide by zero
SolomonZelman
  • SolomonZelman
technically u r...
SolomonZelman
  • SolomonZelman
when you divide by a number that approaches 0 from the right, you get ∞, and when you divide by a number that approaches 0 from the left, you get -∞. if the divident is positive. In this case divident is negative, so it is the other way around. when you divide by a number that approaches 0 from the right, you get -∞, and when you divide by a number that approaches 0 from the left, you get ∞. And if you take a 2-sides limit (as in your case), you shouldn't get just ∞ or -∞.....

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