# Simple limit question. need a little help please

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??

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Yes, because the exponential function, ax, is continuous.

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

I'm not sure what you mean. I was referring to the general fact that, from the definition of "continuous", if xn 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?

Maybe the squeeze theorem?

glenn

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))

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.