Question #6dc23

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.)

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]}