roselin
whats the lim x goes to infinity of 2/square root of x
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JoãoVitorMC
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roselin
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yes
zepdrix
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Rose if you're not convinced, just plug a large number into your calculator to check! :)
if you plug one bajillion in for x, what value do you get for the limit?
\[\frac{ 2 }{ \sqrt{bajillion} }\approx \frac{ 2 }{ 999999999 } \approx 0\]
JoãoVitorMC
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yeah, but do what zepdrix said only for check, because if you do this in a test for example, it will be wrong because infinity is NOT a number...
roselin
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okay
roselin
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it is always 0 ?
JoãoVitorMC
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no! it will depends on what you are taking the limit, always try to simplify the limit, try to do not have a denominator, in this case we can not simplify so we do this
roselin
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i am trying to solve this problem where x goes to infinity 2+square root of x/ 2- square root of x
roselin
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i am getting the answer as 1, is it right?
JoãoVitorMC
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\[\lim_{x \rightarrow \infty}\frac{2+\sqrt{x}}{2-\sqrt{x}}\]
JoãoVitorMC
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this one?
roselin
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yes,
JoãoVitorMC
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the limit is equal to -1
JoãoVitorMC
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have you already study L'hospital rule ?
roselin
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can you please explain how you got -1?
roselin
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No,
JoãoVitorMC
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ok let me explain
JoãoVitorMC
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|dw:1350703643878:dw|
JoãoVitorMC
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got it?
roselin
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what did you do with the whole numerator thing? u ended up with X
JoãoVitorMC
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when x goes to infinity, the other terms with lower power we take them off, and stay only with the higher terms them simply if possible and aply the limit
JoãoVitorMC
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i will post how i got the numerator
roselin
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ohh , u mean the limit of 4 and 4 sqrt x = 0?
JoãoVitorMC
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|dw:1350704041192:dw|
JoãoVitorMC
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no this i simply take them out because comparing with infinity they make no difference
roselin
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oh okay,
JoãoVitorMC
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when you deal with limits on the infinity you can make this assumptions
JoãoVitorMC
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but take care...
JoãoVitorMC
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when you see L'hostpital's Rule this limits will became very easy to solve.... you will see
roselin
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we have not gotten to that yet.
JoãoVitorMC
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yeah
roselin
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if you do not mind then, I have one more problem. would you explain that one?
JoãoVitorMC
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sure
roselin
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thank you
roselin
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lim sec x
x goes to( -pi/2)^+
JoãoVitorMC
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let me solve here one moment
roselin
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okay
JoãoVitorMC
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it is equal to infinity, i will draw here
roselin
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okay
JoãoVitorMC
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|dw:1350704691784:dw|
JoãoVitorMC
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is it clear?
roselin
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yes,viewing it.
JoãoVitorMC
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see, when we have something divide by a number really really really small them it becomes a really really really large number so the limit tends to the infinity
roselin
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okay, but how did you get that 10?
JoãoVitorMC
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it is just a notation for this: 0.00000000000000000000000...00000000000001, then we represent this really small or large number with 10 power
roselin
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oh ,okay. got it.
roselin
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wait,wait
roselin
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that (-pi/2)^+ is the same as (-pi/2)?
JoãoVitorMC
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ALMOST the same, when we have this + is a little more than pi/2
JoãoVitorMC
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-pi/2, sorry
roselin
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so, the asnwer will be the same one right?
roselin
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*answer
JoãoVitorMC
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no, because if we take 1/0 it is not allowed. But 1/0.00000000001 is allowed
JoãoVitorMC
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we can not divide by 0, NEVER
roselin
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yes,that is true.
JoãoVitorMC
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so you got it?
roselin
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yeah.
roselin
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Thank you soooo much.
JoãoVitorMC
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yw :)
roselin
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good night, byee
JoãoVitorMC
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good, bye