General

no worries, see other thread
/thread

@bloodIsShed
@Shimy
Oh my god I am so sorry. I was supposed to put a division sign between the two parentheses, so it isn't multiplication. I truly apologize if this took you a lot of time.

trying to show something that is false to be true is an exercise in futility
either the TS copied the expression incorrectly (maybe missed a negative sign somewhere?), or the answer in the book is wrong (rare, but it happens. if the author knows about this, he/she probably put a note in the errata section in her/his website). of course, it's also possible that the 'answer' of e^-3 belongs to a completely different problem.

welp i tried
Wolfram op.

[quote=Shimy]It should be noted that e = (lim x -> infinity) (1+1/x)^x. We can rewrite x as n = 1/x when x approaches infinity n = 0
(lim n to 0) (1+n)^(1/n)
So now we a form of (1+1/x) to find our n

Lets break up the brackets into
(x-2)(x+1) = (x^2 -x - 2)
In order to get that form we need to get rid of our x variables
(x^2/x^2 - x/x^2 - 2/x^2) -> (1 - 1/x - 2/x^2) *note it doesn't matter if it's 1/x^2 or 1/x^200 its the same because of the limit as x -> infinity
Rewriting it :
(1-3/x)^x

Okay this is where it might get a bit confusing.
n = -3/x. so x = n / -3
Plug in n/-3 for your x in your equation
You'll get (lim n -> 0) (1+n)^(-3/n)
This can be rewritten as :
(lim n -> 0)(1+n^(1/n))^-3 *using basic exponent rules
Or
e^-3

This question's really hard. If it asks to evaluate using the limit definition. Just cry.[/quote]

it's all nice and well except for two problems:

lim (x^2 -x -2)^x as x approaches infinity is NOT the same as lim (x^2/x^2 -x/x^2 -2/x^2)^x as x approaches infinity
consider this: is lim x as x approaches infinity the same as lim x/x^2 as x approaches infinity? obviously not.

the second problem is near the end
lim (1 -1/x - 2/x^2)^x as x approaches infinity is NOT the same as lim (1 - 3/x)^x as x approaches infinity
lim (1 -1/x - 2/x^2)^x as x approaches infinity = 1/e
lim (1 - 3/x)^x as x approaches infinity = 1/e^3
(neither of these are correct, as the solution is [url=http://www.wolframalpha.com/input/?i=lim+%28x^2+-x+-2%29^x+as+x+approaches+infinity]infinity[/url]

It should be noted that e = (lim x -> infinity) (1+1/x)^x. We can rewrite x as n = 1/x when x approaches infinity n = 0
(lim n to 0) (1+n)^(1/n)
So now we a form of (1+1/x) to find our n

Lets break up the brackets into
(x-2)(x+1) = (x^2 -x - 2)
In order to get that form we need to get rid of our x variables
(x^2/x^2 - x/x^2 - 2/x^2) -> (1 - 1/x - 2/x^2) *note it doesn't matter if it's 1/x^2 or 1/x^200 its the same because of the limit as x -> infinity
Rewriting it :
(1-3/x)^x

Okay this is where it might get a bit confusing.
n = -3/x. so x = n / -3
Plug in n/-3 for your x in your equation
You'll get (lim n -> 0) (1+n)^(-3/n)
This can be rewritten as :
(lim n -> 0)(1+n^(1/n))^-3 *using basic exponent rules
Or
e^-3

This question's really hard. If it asks to evaluate using the limit definition. Just cry.

are you sure you didn't make a mistake somewhere, because the way it is written, the value should increase indefinitely as x increases.
(that is, the limit would be infinity, which is to say that it does not converge to a single value)