whats the lim x goes to infinity of 2/square root of x

- anonymous

whats the lim x goes to infinity of 2/square root of x

- schrodinger

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- anonymous

0

- anonymous

yes

- zepdrix

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?
\[\frac{ 2 }{ \sqrt{bajillion} }\approx \frac{ 2 }{ 999999999 } \approx 0\]

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## More answers

- anonymous

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...

- anonymous

okay

- anonymous

it is always 0 ?

- anonymous

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

- anonymous

i am trying to solve this problem where x goes to infinity 2+square root of x/ 2- square root of x

- anonymous

i am getting the answer as 1, is it right?

- anonymous

\[\lim_{x \rightarrow \infty}\frac{2+\sqrt{x}}{2-\sqrt{x}}\]

- anonymous

this one?

- anonymous

yes,

- anonymous

the limit is equal to -1

- anonymous

have you already study L'hospital rule ?

- anonymous

can you please explain how you got -1?

- anonymous

No,

- anonymous

ok let me explain

- anonymous

|dw:1350703643878:dw|

- anonymous

got it?

- anonymous

what did you do with the whole numerator thing? u ended up with X

- anonymous

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

- anonymous

i will post how i got the numerator

- anonymous

ohh , u mean the limit of 4 and 4 sqrt x = 0?

- anonymous

|dw:1350704041192:dw|

- anonymous

no this i simply take them out because comparing with infinity they make no difference

- anonymous

oh okay,

- anonymous

when you deal with limits on the infinity you can make this assumptions

- anonymous

but take care...

- anonymous

when you see L'hostpital's Rule this limits will became very easy to solve.... you will see

- anonymous

we have not gotten to that yet.

- anonymous

yeah

- anonymous

if you do not mind then, I have one more problem. would you explain that one?

- anonymous

sure

- anonymous

thank you

- anonymous

lim sec x
x goes to( -pi/2)^+

- anonymous

let me solve here one moment

- anonymous

okay

- anonymous

it is equal to infinity, i will draw here

- anonymous

okay

- anonymous

|dw:1350704691784:dw|

- anonymous

is it clear?

- anonymous

yes,viewing it.

- anonymous

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

- anonymous

okay, but how did you get that 10?

- anonymous

it is just a notation for this: 0.00000000000000000000000...00000000000001, then we represent this really small or large number with 10 power

- anonymous

oh ,okay. got it.

- anonymous

wait,wait

- anonymous

that (-pi/2)^+ is the same as (-pi/2)?

- anonymous

ALMOST the same, when we have this + is a little more than pi/2

- anonymous

-pi/2, sorry

- anonymous

so, the asnwer will be the same one right?

- anonymous

*answer

- anonymous

no, because if we take 1/0 it is not allowed. But 1/0.00000000001 is allowed

- anonymous

we can not divide by 0, NEVER

- anonymous

yes,that is true.

- anonymous

so you got it?

- anonymous

yeah.

- anonymous

Thank you soooo much.

- anonymous

yw :)

- anonymous

good night, byee

- anonymous

good, bye

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