If we multiply 1 by itself any number of times then we get 1 but if we consider 1 raised to the power infinity then it's undefined. This has baffled me and I am unable to understand this case even I know that infinity is not a number. I need a satisfactory answer.
Asked
Active
Viewed 80 times
-1
-
http://math.stackexchange.com/questions/10490/why-is-1-infty-considered-to-be-an-indeterminate-form There are some detailed answers here. – Tom Tseng Jul 11 '15 at 20:08
2 Answers
0
You have answered your own question. Infinity is not just a number. So raising 1 to power infinity doesnt make sense(it is undefined). If it does, then you might as well raise 1 to power "house" or "car" or anything. That is the meaning of undefined. However, $\lim\limits_{n\rightarrow\infty}1^n=1.$
OKPALA MMADUABUCHI
- 1,063
- 6
- 16
0
Perhaps you are misremembering the statement that $1^\infty$ is an indeterminate form. That is, from only the information $$ \lim_{t \to a} f(t) = 1,\qquad\text{and}\qquad \lim_{t \to a} g(t) = \infty $$ it is not possible to determine $$ \lim_{t \to a} f(t)^{g(t)} . $$
For example, $$ \lim_{t \to 0} (1+t^2) = 1, \qquad \lim_{t \to 0} \frac{1}{t^2} = \infty \qquad \lim_{t \to 0} (1+t^2)^{1/t^2} = e \\ \lim_{t \to 0} (1+t) = 1, \qquad \lim_{t \to 0} \frac{1}{t^2} = \infty \qquad \lim_{t \to 0} (1+t)^{1/t^2} = \infty \\ \lim_{t \to 0} (1+t^2) = 1, \qquad \lim_{t \to 0} \frac{1}{t} = \infty \qquad \lim_{t \to 0} (1+t^2)^{1/t} = 1 $$
GEdgar
- 111,679