- #1
Damidami
- 94
- 0
Let [tex]\{a_n\}[/tex] be a sequence of positive numbers. If [tex]\lim a_n = A[/tex], prove that [tex]\lim \sqrt{a_n} = \sqrt{A}[/tex].
I thoght that [tex]f(x) = \sqrt{x}[/tex] is a continuous function, and composing a sequence that has a limit with a continuous function must have a limit that is just evaluating the previous limit in the function.
But I think I'm not supposed to use that fact, as the concept of limit of sequence comes previously to the definition of continuous function in the course.
It seems trivial, but still can't find how to start. Any help?
I thoght that [tex]f(x) = \sqrt{x}[/tex] is a continuous function, and composing a sequence that has a limit with a continuous function must have a limit that is just evaluating the previous limit in the function.
But I think I'm not supposed to use that fact, as the concept of limit of sequence comes previously to the definition of continuous function in the course.
It seems trivial, but still can't find how to start. Any help?