infinity limits

[racuna]

Find limit f(x) as x->infinity if, for all x>1, (4e^x-24)/(2e^x)<f(x)<(2sqrt(x))/(sqrt(x)-1)
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Find a formula for a function f that satisfies the following conditions:
lim f(x) x->infinity=0, lim f(x) x->0=-infinity, f(2)=0, lim f(x) x->3-=infinity, and lim f(x) x->3+=-infinity

It says to fill in the blanks:

f(x)=(__-x)/(x^_(x-__))
I have no idea how to even start these problems. Please help!
-Rachel

 

[tkhunny]

Find limit f(x) as x->infinity if, for all x>1, (4e^x-24)/(2e^x)<f(x)<(2sqrt(x))/(sqrt(x)-1)

Find this

limit as x->infinity for (4e^x-24)/(2e^x)

and find this

limit as x->infinity for (2sqrt(x))/(sqrt(x)-1)

It is hoped that they are the same. If so, we know the same limit for f(x) is the same, since it is SQUEEZED between them. If they are not the same, we have quite an inconclusive result.

 

[pka]

Your problem is also algebra. It is hard to do calculus without basic algebra.
[(4e<SUP>x</SUP>−24)/(2e<SUP>x</SUP>)=[2−(12/e<SUP>x</SUP>]
Moreover, 2√(x)/(√(x)−1)= 2/(1−[1/(√(x)]

 

[racuna]

How do you compute the second problem I posted? (Below the line)

 

[soroban]

Hello, Rachel!

Find a formula for a function f(x) that satisfies the following conditions:
lim(x->inf) f(x) = 0, lim(x->0) f(x) = - infinity, f(2) = 0,
lim(x->3-) f(x) = infinity, and . lim(x->3+) f(x) = -infinity

It says to fill in the blanks:

f(x) = (__ - x)/ [x^_ (x - __)]

With that skeleton form, they've practically given away the answer.

Since f(2) = 0, the numerator must have: 2 - x.

To have a vertical asymptote at x = 3, there must be (x - 3) in the denominator.

To have a vertical asymptote at x = 0, there must be x in the denominator.
. . Since we want the function to go to negative infinity from both sides,
. . the denominator must contain x².

. . . . . . . . . . . . . . . . . . . . . . 2 - x
The function is: . f(x) . = . ------------
. . . . . . . . . . . . . . . . . . . . .x²(x - 3)