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yes

okay

it is always 0 ?

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

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

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

this one?

yes,

the limit is equal to -1

have you already study L'hospital rule ?

can you please explain how you got -1?

No,

ok let me explain

|dw:1350703643878:dw|

got it?

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

i will post how i got the numerator

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

|dw:1350704041192:dw|

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

oh okay,

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

but take care...

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

we have not gotten to that yet.

yeah

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

sure

thank you

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

let me solve here one moment

okay

it is equal to infinity, i will draw here

okay

|dw:1350704691784:dw|

is it clear?

yes,viewing it.

okay, but how did you get that 10?

oh ,okay. got it.

wait,wait

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

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

-pi/2, sorry

so, the asnwer will be the same one right?

*answer

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

we can not divide by 0, NEVER

yes,that is true.

so you got it?

yeah.

Thank you soooo much.

yw :)

good night, byee

good, bye