What is the limit of g'(x) on the interval (a, ∞) as x approaches infinity?

Sethka
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When a question asks along the lines of :
"If a function (g) is decrasing on the interval {x,x)...What would the limg'(x) be (As it approaches infinity)"

What are they looking for? and whatequation am I using? I'm not looking for too much info on how to do, but which direction should I go in? I'm not sure what topic this would lie in.
 
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If g(x) is decreasing on the interval, say x\in \[ a,\infty) then, unless you know more about the function g(x), all you can say is \lim_{x\rightarrow\infty}g^{\prime}(x)<0, at least that I can tell.
 
"decreasing on the interval {x, x)" makes no sense. In any case, it order to talk about a limit at infinity, we would have to know what happens on some unbounded interval, say (a, \infty) as benorin said. And you still can't answer the question except as he said.

For example, if g(x)= ax, with a any negative number, then g'(x)= a for all x and so has limit a. Without more information about g, the limit could be any negative number.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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