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For a given function #f(x)=x^a/x^b#, is it correct to say that #f(0)=0,1or"undefined"# depending on the values of #a# and #b#?

For example, take #f(x)=x^2/x^2#, we can just simplify this to #f(x)=1#, and say #f(0)=1#, however we can also say that #f(0)=0^2/0^2=0/0="undefined"#

Or, with #f(x)=x^2/x#, we can simplify this to #f(x)=x# and so #f(0)=0#. However, we can also make it so #f(0)=0^2/0="undefined"#.

Answer 1

Cameron Alexander

Think about this:

#f(x) =x^a/x^b = x^(a-b) =x^c#

with #c=a-b#. So:

#a=b => c=0 => f(0) = 1#

#a > b => c > 0 => f(0) = 0#

#a < b => c < 0 => f(0)# is undefined

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Answer 2

Adam Adkins

For the given function ( f(x) = \frac{x^a}{x^b} ), the value of ( f(0) ) depends on the exponents ( a ) and ( b ):

If ( a > b ), ( f(0) = 0 ).
If ( a = b ), ( f(0) = 1 ).
If ( a < b ), ( f(0) ) is undefined due to division by zero.

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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.

For a given function #f(x)=x^a/x^b#, is it correct to say that #f(0)=0,1or"undefined"# depending on the values of #a# and #b#?

For example, take #f(x)=x^2/x^2#, we can just simplify this to #f(x)=1#, and say #f(0)=1#, however we can also say that #f(0)=0^2/0^2=0/0="undefined"# Or, with #f(x)=x^2/x#, we can simplify this to #f(x)=x# and so #f(0)=0#. However, we can also make it so #f(0)=0^2/0="undefined"#.

For example, take #f(x)=x^2/x^2#, we can just simplify this to #f(x)=1#, and say #f(0)=1#, however we can also say that #f(0)=0^2/0^2=0/0="undefined"#

Or, with #f(x)=x^2/x#, we can simplify this to #f(x)=x# and so #f(0)=0#. However, we can also make it so #f(0)=0^2/0="undefined"#.