Is the Limit of Powers Equal to the Power of Limits?

[sutupidmath]

i need someone to coment on this:

lim(a^x_n)^x'_n, when n->infinity = (a^lim x_n)^lim x'_n , n-> infinity,

what i am asking here is if we can go from the first to the second? or if these two expressions are equal??

any help??

 

[HallsofIvy]

Yes, because the exponential function, a^x, is continuous.

 

[sutupidmath]

one more thing here. Is there a theorem or a deffinition that supports similar expressions in a more generalized way?? i forgot to mention this also

 

[HallsofIvy]

I'm not sure what you mean. I was referring to the general fact that, from the definition of "continuous", if x_n is a sequence of numbers converging to a and f is a function continuous at a, then
\lim_{n \rightarrow \infnty} f(x_n)= f(\lim_{n\rightarrow \infty} x_n)= f(a)

 

[sutupidmath]

I'm not sure what you mean. I was referring to the general fact that, from the definition of "continuous", if x_n is a sequence of numbers converging to a and f is a function continuous at a, then
\lim_{n \rightarrow \infnty} f(x_n)= f(\lim_{n\rightarrow \infty} x_n)= f(a)

yeah this is what i am asking. But what i want to know is if there is a theorem that states this, what you wrote. Or how do we know that this is so?

 

[cybercrypt13]

Maybe the squeeze theorem?

glenn

 

[Data]

As Halls indicated, it's essentially the definition of continuity. The definition of continuity for a function f of one real variable defined on an interval (a,b) is for any x in (a,b),

\lim_{y \rightarrow x} f(y) = f(x)

(ie. the limit exists and is equal to f(x))

 

[sutupidmath]

As Halls indicated, it's essentially the definition of continuity. The definition of continuity for a function f of one real variable defined on an interval (a,b) is for any x in (a,b),

\lim_{y \rightarrow x} f(y) = f(x)

(ie. the limit exists and is equal to f(x))

Yeah, i know the definition of continuity, i was just wondering if there is a specific theorem that states this, as i have not encountered one on my calculus book. However, i do understand it now.
Many thanks to all of you.