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October 21, 2014

October 21, 2014

Posted by **maura (urgent!!)** on Wednesday, September 12, 2012 at 9:53pm.

1) To evaluate

lim x→infinity sqrt (x^2 + 4)

first consider that as x becomes infinitely large, x^2+4 -> _____

2)Find lim x→ infinity f(x)if,

for all x > 1,

(6e^x − 17)/3e^x < f(x) < 2sqrt(x)/ sqrt (x − 1)

- calculus -
**maura (urgent!!)**, Wednesday, September 12, 2012 at 10:09pmforget the first one i got it. just the second one i dont know where to begin

- calculus -
**Reiny**, Wednesday, September 12, 2012 at 10:42pmfor (6e^x - 17)/(3e^x)

divide top and bottom by e^x

we get

(6-17/e^x)/3 , as x ----> ∞ , the expression ---> 2

and 2√x/√(x-1) ----> 2 , as x ---->∞

so as x---->∞

2 < f(x) < 2

which implies f(x) = 2

- calculus -
**maura (urgent!!)**, Wednesday, September 12, 2012 at 10:48pmSOOOO HAPPY YOU ANSWERED ME!

ok now im just trying to process this in my head. i understand dividing it by e^x but you lost me at the expression -->2

sorry to be a pain in the ass, i just want to make sure i fully understand this. thank you so much! you are a life saver!

- calculus -
**Reiny**, Wednesday, September 12, 2012 at 11:00pmin (6-17/e^x)/3 , as x ----> ∞

what happens when you divide 17 by a very very very large number ?

Doesn't it go to zero ?

so you have (6-0)/3 = 2

for the right side 2√x/√(x-1)

when x is really really large, do you think it makes any difference if we take √x or √(x-1) ?

e.g. take

√123456789 and √123456788

now divide them to get 1.000000004

so the right side would be 2.00000008 or just plain 2

so the f(x) is sandwiched between 2 and 2 , thus it MUST BE 2

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