Question

2 raise to power infinity=?

Answer 1

infinity

Answer 2

Well, 2^2 =4
2^5 = 32
2^10 = 1024

So it keep on growing faster and faster (exponentially!), so as

n --> infinity, 2^n --> infinity, also.

Answer 3

infinity. x raise power infinity = infinity (if x!=0)

Answer 4

fun side note...
2^inf => inf much quicker than inf => inf.

Answer 5

@ajakum... : actually only if x > 1

Answer 6

u mean x raise to power is infinity=infinity and a constant raise to power infinity = infinity

Answer 7

\[x^\infty \rightarrow \infty \iff x > 0\]

Answer 8

and if x<0 then=?

Answer 9

indeterminate

Answer 10

\[x^{\infty} \to \infty \:\:\: when \:\:\:x > 1 \]

Answer 11

\[x^{\infty} \to 0 , 0<x<1\]

Answer 12

\[x^{\infty} =1, x=1\]

Answer 13

-x ^ inf -> -inf iff x < -1
and before I should have said x^inf -> inf iff x > 1
I forgot about the indeterminace

Answer 14

@agreene: false.

Consider x = 1/2

Then\[\lim_{n \rightarrow \infty} x^{n} = \lim_{n \rightarrow \infty} (1/2)^{n} = 0.\]

In fact, this is true for all x such that 0 < x < 1.

When x = 1, clearly the limit is 1 and when x = 0 the limit is 0.

For x such that -1 < x < 0, the limit is zero

But for x =< -1, there is no limit as the sign keep on alternating:
e.g., x = -2
x^1 = -2, x^2 = 4, x^3 = -8, x^4 = 16, x^5 = -32

Answer 15

@JamesJ
yes, thanks for this clarity. I realized that in my haste I had overlooked it not long after the next question was asked.

Answer 16

u mean when -1 <x>1 then x^infinity = 0 when x=1 then x^infinity=1 when x=0 then x^infinity=0 and when x(is less than and =)-1 then x^infinity=does not exist. and when x>1 then x^infinity=infinity am i right???

Answer 17

you are close.
x^inf -> inf if x > 1
x^inf -> - inf if x < -1
x^inf = 0 iff x = 0
x^inf = indeterminate iff -1 < x < 1 and x =/= 0

Answer 18

In summary:
\[ lim_{n \rightarrow \infty} \ x^n \ = \left\{ \begin{array}{ll} \infty & x > 1 \\ 1 & x = 1 \\ 0 & 0 \leq x < 1 \\ 0 & -1 < x < 0 \\ n.d. & x \leq -1 \end{array} \right. \]

Answer 19

thank u soo much james u have solved a major problem of me.