what is lim(x/lnx)+2 when x tends to infinity???

Answer 1

lim(x/ln(x)) + 2
x -> infinity

To solve this, we use L'Hospital's rule, as we have the form
[infinity/infinity]. Therefore, taking the derivative of the top and bottom,

lim (1/(1/x)) + 2
x -> infinity

=

lim (x) + 2
x -> infinity

Which doesn't exist.

Answer 2

an easy way to figure to see what is growing fastest, x or lnx? if x is fastest then it will be infinity but if lnx is faster then the first term will be zero. forget about the +2 for now. in calculus you learn a method of L'hopitals. where you take the derivative of the top divided by the derivative of the bottom thus it would turn out 1/(1/x) as x goes to infinity the term would get huge!! thus it goes to infinity.

Answer 3

Use L'Hospital's rule for indeterminant forms. In this case it is ∞/∞.

lim x→∞ of [(x/lnx)] + 2 = lim x→∞ of [1/(1/x)] + 2 = 1/0 + 2 = ∞

Answer 4

Well, sounds like you would use L'hopitals rule since when you plug in infinity for x you get infinity divided by infinity.

Therefor you must take the derivitive of the top divided by the derivitive of the bottom and you will get:

1/(1/X) = X

Lim (x), x-> infinity = Infinity

Answer 5

It's opposite to a decaying exponential which keeps going to zilch.

But the answer is "a little more than" infinity. Not a lot more mind you, but just a little more... so don't exaggerate it.

Right now I'm writing it out in decimal form so it will be a while before I get back with you... a little longer than eternity but not a lot more.

BRB

Answer 6

Puggy is right...I does not exist...or in other words it is infinity

Answer 7

Is this an actual question because you WANT to know or is this your homework?

Go to math.com

Answer 8

∞