Question

Question #6dc23

Answer 1

For mathematics at calculus and beyond, we leave it undefined.

Explanation:

I am not aware of any algebraic difficulties in defining `1^oo` to be `1`. Although there are some challenges in trying to treat infinity like a number in the regular number system. You may find it interesting to look into the "extended real numbers" or the "extended real number system". (You also may learn a lot of mathematics)

For me, the problem arises when we look at the behavior of certain functions (algebraic expressions).

As `x` increases without any bound on how big it gets, the fraction `1/x` gets closer and closer to `0`.

So the expression `1+1/x` gets closer and closer to `1`.

Now look at these two graphs:

The first is for `(1+1/x)^(2x)`

As `x` gets bigger and bigger, `1+1/x` gets close to `1` and the exponent, `2x` gets bigger and bigger (Like infinity?). However `y` does NOT get closer to `1`, it gets closer and closer to `7.38`

graph{(1+1/x)^(2x) [-1.75, 30.28, -3.75, 12.27]}

Here is another graph. This one is for `(1-1/x)^x`

Again, as `x` gets bigger and bigger, `y` dos not get closer and closer to `1`, it gets closer to `0.368`

graph{(1-1/x)^x [-4.08, 11.72, -2.784, 5.12]}

When we start looking at "limits" we want to leave `1^oo` undefined because of expressions like these.

(There are other reasons, but I think this is easiest to explain -- because we can look at pictures.)

Answer 2

There is one sense in which we can think of `1^(infty)=1`: through the idea of an infinite product . As Jim mentions, in most other situations, `1^(infty)` is called an "indeterminate form " and can take on various values.

Explanation:

An infinite product is an expression of the form `Pi_{n=1}^{infty}a_{n}=a_{1}*a_{2}*a_{3}* * *` (the capital "pi" represents "product" and is analogous to a capital "sigma" for summation notation). Such a product is said to converge if the sequence of partial products, defined by `p_{n}=a_{1}*a_{2}*a_{3}*...*a_{n}`, converges.

For example, if `a_{n}=e^(1/2^{n})`, then `p_{n}=e^(1/2) * e^(1/4) * e^(1/8) * * * e^(1/2^{n})=e^{1/2+1/4+1/8+ * * * + 1/2^(n)}=e^{1-1/2^(n)}` (look up facts about geometric series to see where this last equality comes from ).

Since `p_{n}->e` as `n->infty`, the corresponding infinite product converges and we can write `Pi_{n=1}^{infty}e^(1/2^{n})=e`.

If we therefore interpret `1^{infty}` as an infinite product `1*1*1*1* * *`, then `p_{n}=1` for all `n` and it makes sense to say `1^{infty}=1`.

In most other situations, the expression `1^{infty}` arises when considering a function of the form `h(x)=f(x)^(g(x))`, where, as `x` approaches some "value" (like zero or infinity), the value of `f(x)` is getting closer and closer to 1 while the value of `g(x)` is approaching `infty` (it's getting "arbitrarily large"). This is one example of an "indeterminate form " in mathematics.

In Calculus, a technique called L'Hopital's Rule is often used to try to evaluate what the value of `h(x)` is approaching as `x` gets closer and closer to the relevant value. In the present situation, it is also helpful to take the natural logarithm of both sides of the equation `h(x)=f(x)^(g(x))` to get `ln(h(x))=g(x)ln(f(x))` (after using a property of logarithms). This is then typically rewritten in the form `ln(h(x))=(ln(f(x)))/(1/g(x))` before L'Hopital's Rule is used.

If, for example, `ln(h(x))=(ln(f(x)))/(1/g(x))` gets closer and closer to 1 as `x` gets closer and closer to 0, it follows that `h(x)` itself is getting closer and closer to `e^{1}=e` as `x` gets closer and closer to zero.

This is similar to what is happening in Jim's example. If we modify the original function to be `h(x)=(1+x)^(1/x)` and let `x->0` (rather than `x->infty`). Here, `f(x)=1+x` and `g(x)=1/x`. It follows that `ln(h(x))=(ln(1+x))/x` and you can see from the graph of `(ln(1+x))/x` below that it is approaching 1 as `x` gets closer and closer to zero. Therefore, `h(x)=(1+x)^(1/x)` gets closer and closer to `e` as `x` gets closer and closer to zero. The graph of `h(x)` is shown even further below.

graph{ln(1+x)/x [-5, 5, -2.5, 2.5]}

graph{(1+x)^(1/x) [-10, 10, -5, 5]}