Find the limit of the following function as x goes to infinity?

Question

f(x) = 520*sqrt { x * ( [2 / (x+1)] ^ [ (x+1) / (x-1) ] ) }.
What is lim f(x) as x ----> infinity.

Answer 1

f(x) = 520*sqrt { x * ( [2 / (x+1)] ^ [ (x+1) / (x-1) ] ) }.

First off, we're going to do synthetic long division on (x+1)/(x-1). Without showing the steps involved, the answer would be:
2/(x - 1) + 1

f(x) = 520*sqrt { x * ( [2/(x+1)]^[2/(x-1) + 1] ) }

Now, we can effectively split [2/(x+1)]^[2/(x-1) + 1] into
[2/(x+1)]^[2/(x-1)] * [2/(x+1)]^1. So we now have

f(x) = 520*sqrt {x * [2/(x+1)] ( [2/(x+1)]^[2/(x-1)] ) }

And, with that done, we can combine the first two terms in the { } brackets.

f(x) = 520 * sqrt { [2x/(x+1)] ( [2/(x+1)]^[2/(x-1)] ) }

Not only that, but we can take the square root of each term in the brackets. Note that now there are only a product of two expressions.

f(x) = 520 * sqrt( [2x/(x+1)]) * sqrt ( [2/(x+1)]^[2/(x-1)] )

Notice that taking the square root of something is the same as taking it to the power of 1/2. That means for our second square root, we'll be taking a power to a power, which means we multiply the exponents (i.e. (a^b)^c = a^(bc) ). This means our final exponent will be (1/2) * (2/(x - 1)) = (1/(x-1)).

f(x) = 520 * sqrt ( [2x/(x+1)]) * [2/(x+1)]^[1/(x - 1)]

Keep in mind we haven't even taken the limit yet. We can, however, solve this in terms of individual limits. To avoid confusion, I'm going to assign

f(x) = 520 * g(x) * h(x), where
g(x) = sqrt ( [2x/(x+1)]) and
h(x) = [2/(x+1)]^[1/(x - 1)]

Let's solve lim (x -> infinity) g(x) first.

lim { sqrt ( [2x/(x+1)]) } = sqrt (lim [2x/(x+1)])
x -> infinity
= sqrt (lim [2/(1 + 1/x)]) = sqrt (2/(1 + 0))
= sqrt(2)

Now, let's solve lim (x -> infinity, h(x)).

lim [2/(x+1)]^[1/(x - 1)]
x -> infinity

Here, we get the form (0^0), which is an indeterminate form. Recall that [f(x)]^[g(x)] = e^[g(x) ln f(x) ].
I'm not going to rewrite (lim x -> infinity in the next few steps).

e^( [1/(x-1)] * ln (2/(x+1)) ) =
e^ ( ln (2/(x+1)) / (x - 1) )

Applying the log property for subtraction,

e^ ( [ln2 - ln(x+1)] / (x - 1) )

In our exponent, we have the form [infinity/infinity], so we use L'Hospital's rule, to obtain

e^ ( [-1/(x+1)] / (1) ) = e^ ( [-1/(x+1)] )

Now, we take the limit as x approaches infinity, and our final answer is e^0 = 1.

Back to our original question,
lim f(x) = 520 * sqrt(2) * 1 = 520*sqrt(2)
x -> infinity

Answer 2

Start with the part (x + 1) / (x - 1)

Divide the numerator by x and
divide the denominator by x, to get :
(1 + 1/x) / (1 - 1/x)

As x tends to infinity, 1/x tends to zero.
This last expression then becomes
(1 + 0) / (1 - 0) = 1 / 1 = 1.
That is, the exponent tends to 1.

Now, f(x) = 520 * sqrt [ x * 2 / (x + 1) ]
= 520 * sqrt [ 2x / (x + 1)]

Now do much the same with the part 2x / (x + 1).
Dividing each by x gives :
2 / (1 + 1/x)

Again, 1/x tends to zero, so the expression becomes :
2 / (1 + 0) = 2 / 1 = 2

Finally then, f(x) = 520 * sqrt(2), as x tends to infinity.

Answer 3

f(x) = 520*sqrt { x * ( [2 / (x+1)] ^ [ (x+1) / (x-1) ] ) }.
lim x→∞ f(x) = lim x→∞ 520 √{x * [2 / (x+1)] ^ 1
= lim x→∞ 520 √[2x / (x+1)] = 520√2

Answer 4

As n is going to infinity, ln n is going to infinity. yet as n is going to infinity, x^n is going to infinity if x > 0, maintains to be as 0 if x = 0, and has no limits if x < 0. Then, the decrease is undefined for x < 0, is 0 for x = 0 and infinity for x > 0.

Answer 5

Zero.

lim as x---> infinity of 2/(x+1) is zero. Zero raised to any power is still zero, as is it's square root. Multiplying by 520 still leaves zero.

Answer 6

just by looking at it,, it either converges to 520 or 0

Answer 7

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