# Question #6dc23

Dec 18, 2016

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

#### Explanation:

I am not aware of any algebraic difficulties in defining ${1}^{\infty}$ 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 $\frac{1}{x}$ gets closer and closer to $0$.

So the expression $1 + \frac{1}{x}$ gets closer and closer to $1$.

Now look at these two graphs:

The first is for ${\left(1 + \frac{1}{x}\right)}^{2 x}$

As $x$ gets bigger and bigger, $1 + \frac{1}{x}$ gets close to $1$ and the exponent, $2 x$ 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 ${\left(1 - \frac{1}{x}\right)}^{x}$

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 ${1}^{\infty}$ undefined because of expressions like these.

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

Dec 20, 2016

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 ${\Pi}_{n = 1}^{\infty} {a}_{n} = {a}_{1} \cdot {a}_{2} \cdot {a}_{3} \cdot \cdot \cdot$ (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 ${p}_{n} = {a}_{1} \cdot {a}_{2} \cdot {a}_{3} \cdot \ldots \cdot {a}_{n}$, converges.

For example, if ${a}_{n} = {e}^{\frac{1}{2} ^ \left\{n\right\}}$, then ${p}_{n} = {e}^{\frac{1}{2}} \cdot {e}^{\frac{1}{4}} \cdot {e}^{\frac{1}{8}} \cdot \cdot \cdot {e}^{\frac{1}{2} ^ \left\{n\right\}} = {e}^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdot \cdot \cdot + \frac{1}{2} ^ \left(n\right)} = {e}^{1 - \frac{1}{2} ^ \left(n\right)}$ (look up facts about geometric series to see where this last equality comes from ).

Since ${p}_{n} \to e$ as $n \to \infty$, the corresponding infinite product converges and we can write ${\Pi}_{n = 1}^{\infty} {e}^{\frac{1}{2} ^ \left\{n\right\}} = e$.

If we therefore interpret ${1}^{\infty}$ as an infinite product $1 \cdot 1 \cdot 1 \cdot 1 \cdot \cdot \cdot$, 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 $h \left(x\right) = f {\left(x\right)}^{g \left(x\right)}$, where, as $x$ approaches some "value" (like zero or infinity), the value of $f \left(x\right)$ is getting closer and closer to 1 while the value of $g \left(x\right)$ 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 $h \left(x\right)$ 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 $h \left(x\right) = f {\left(x\right)}^{g \left(x\right)}$ to get $\ln \left(h \left(x\right)\right) = g \left(x\right) \ln \left(f \left(x\right)\right)$ (after using a property of logarithms). This is then typically rewritten in the form $\ln \left(h \left(x\right)\right) = \frac{\ln \left(f \left(x\right)\right)}{\frac{1}{g} \left(x\right)}$ before L'Hopital's Rule is used.

If, for example, $\ln \left(h \left(x\right)\right) = \frac{\ln \left(f \left(x\right)\right)}{\frac{1}{g} \left(x\right)}$ gets closer and closer to 1 as $x$ gets closer and closer to 0, it follows that $h \left(x\right)$ 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 $h \left(x\right) = {\left(1 + x\right)}^{\frac{1}{x}}$ and let $x \to 0$ (rather than $x \to \infty$). Here, $f \left(x\right) = 1 + x$ and $g \left(x\right) = \frac{1}{x}$. It follows that $\ln \left(h \left(x\right)\right) = \frac{\ln \left(1 + x\right)}{x}$ and you can see from the graph of $\frac{\ln \left(1 + x\right)}{x}$ below that it is approaching 1 as $x$ gets closer and closer to zero. Therefore, $h \left(x\right) = {\left(1 + x\right)}^{\frac{1}{x}}$ gets closer and closer to $e$ as $x$ gets closer and closer to zero. The graph of $h \left(x\right)$ 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]}