How do you simplify #5^0#?

Answer 1

Please see below.

While it is easy to understand #5^2# and #5^5#, which are #5xx5# and #5xx5xx5xx5xx5#
it is not easy to understand #5^0#.

So let us try it different way.

What is #5^5-:5^2#?
To answer this let us use division as fraction #5^5/5^2# and this is
#(5xx5xx5xx5xx5)/(5xx5)#
= #(5xx5xx5xxcancel5xxcancel5)/(cancel5xxcancel5)#
= #5^3# and we can express this as #5^((5-2)# i.e.
#5^5-:5^2=5^((5-2)#

In fact similarly we can generalize it to say

#a^m-:a^n=a^m/a^n=a^((m-n)#, which is an identity
What is #a^0# then. Well we can use above identity and can interpret it as
#a^0=a^((m-m)#, but RHS is #a^m-:a^m# or #a^m/a^m# i.e. #1#
Hence for any #a#, we have #a^0=1# and so too for #a=5#
and #5^0=1#
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Answer 2

5^0 simplifies to 1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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