Simple equation that I forgot how to do?

Question

How would you find the limit as x goes to infinity of

1 + (-1)^x
----------------
x

My book says it's 0, but I don't know why. When I look at it, I see 1 - infinity divided by infinity--not sure how to simplify that....

thanks

Answer 1

You are seeing things that aren't there.

(-1)^1 = -1
(-1)^2 = 1
(-1)^3 = -1
(-1)^4 = 1
(-1)^5 = -1
(-1)^6 = 1

Notice how it keeps alternating between -1 and 1. So, the numerator keeps alternating between 0 and 2. Meanwhile, the denominator keeps increasing up to infinity, so either:

0 / infinity = 0
2 / infinity = 0

With x even or odd, the limit is still zero.

Answer 2

The function (1+(-1)^x) is bounded between 0 and 2.

the limit of any definite number A (though it is a complex number at times) divided by x as x goes to infin

lim A/X as x -> infin = 0.

Since this is true for 0 and for 2 and any number between its true for the function.

Answer 3

You could use L'Hopital's rule, but this is even easier than that.

The numerator will oscillate between 0 (when x is negative (-1)^x is -1) and 2 (when x is positive (-1)^x is 1)

Limit as x goes to infinity of 0/x is 0 and limit as x goes to infinity of 2/x is also 0

Therefore the limit is 0.

Answer 4

The numerator, 1 + (-1)^x is going to oscillate back and forth between 0 (for odd x) and 2 (for even x). As x approaches infinity, 0/x and 2/x are both going towards zero, so the overall limit is going to approach zero.