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Let f_n(x) = 1 / (1+x)^n for x in (0,1]. Find the pointwise limit f of the sequence (f_n) on (0,1], and determine if (f_n) converges uniformly on (0,1].

User Schadrac
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Final answer:

The pointwise limit f of the sequence (f_n) on (0,1] is the zero function. The sequence (f_n) does not converge uniformly on (0,1].

Step-by-step explanation:

To find the pointwise limit f of the sequence (f_n) on the interval (0,1], we need to take the limit as n approaches infinity of the function f_n(x) = 1 / (1+x)^n. Let's calculate this limit:

As n approaches infinity, the term (1+x)^n will become very small for any value of x in the interval (0,1]. Therefore, the limit of f_n(x) as n approaches infinity will be 0 for any value of x in the interval (0,1]. Hence, the pointwise limit f of the sequence (f_n) is the zero function on the interval (0,1].

To determine if the sequence (f_n) converges uniformly on the interval (0,1], we need to check if the supremum of the pointwise difference between f_n and f approaches zero as n approaches infinity. However, in this case, as n approaches infinity, the pointwise difference between f_n(x) and f(x) = 0 will always be 1 / (1+x)^n, which does not approach zero for any value of x in the interval (0,1]. Therefore, the sequence (f_n) does not converge uniformly on the interval (0,1].

User Obed Castillo
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