Calculus: why is 1^infinity not just 1?

when the limit of a function, through direct substitution, results in 1 to the power of infinity, you have to use L'Hospital's Rule instead of saying that this is equal to 1.

Why is this?

These forms are called indeterminate because if you replace 1, 0, and
infinity by functions the limits of which are 1, 0, and infinity as
x -> 0, then the limit of the compound function does not exist, in the
sense that the limit depends on which functions you choose.

For an example of this discrepancy for 1^infinity, on the one hand,
take f(x) = 1 and g(x) = 1/x. Then:

lim f(x)^g(x) = lim 1^(1/x) = lim 1 = 1

On the other hand, if we take f(x) = 1 + x and g(x) = 1/x, then:

lim f(x)^g(x) = lim (1 + x)^(1/x) = e = 2.718281828459... > 1

its because 1^anything is 1 and anything to the infinity is infinity. so its both at the same time, so it doesnt work. so use lo's

make it a good day

i think, for example 1^-1 is not 1 its .1 or something, i think, i don't know, i don't have a Calculator on me

#If you have any other info about this subject , Please add it free.#

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