Is this graph defined at x = 0?

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SUMMARY

The function f(x) = (2^x - 1) / x is not defined at x = 0 due to the indeterminate form 0/0. The limit of f(x) as x approaches 0 exists and equals ln(2), indicating a removable discontinuity at this point. Unlike f(x), the function g(x) = (2^x - 1) / x for x ≠ 0, with g(0) defined as ln(2), is continuous across all x. This distinction clarifies that while the graph does not exhibit a spike at x = 0, the limit behavior is well-defined.

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PcumP_Ravenclaw
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Hello everyone,
Is the graph ## (2^x - 1) / x ## defined at x = 0?
When I plot this graph there is no spike at 0 because 0/0 is undefined? Is the computer unable to show this? I am finding the limit as x approaches 0. From the graph limit exits but does f(0) exist?

Sehr Danke.
 
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PcumP_Ravenclaw said:
When I plot this graph there is no spike at 0 because 0/0 is undefined? Is the computer unable to show this?

The function f(x)=1/x is also undefined at x=0, but there's a spike in the graph of that function because the limit as x approaches 0 is infinity (or negative infinity if coming from the negative side of zero). There's no spike in (2x-1)/x because the limit as x approaches zero is not infinity.
 
I would say that it is not a question of the "graph" being defined but the function itself. f(x)= (2^x- 1)/x is "not defined" because "0/0" does not correspond to a number. There is what is usually called a "removable" discontinuity at x= 0. "Removable" because the limit, as x goes to 0, of f(x) exists- it is ln(2). The function "g(x)= (2^x- 1)/x if x is not 0, g(0)= ln(x)" is continuous for all x and is exactly the same as f for all x except 0.
 
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