Prove that if lim (x=> c) f(x) exists, then f is bounded on some neighbourhood c

Q: Prove that if lim (x=> c) f(x) exists, then f is bounded on some neighbourhood c

Let F:(a,?) ?R is such that lim xF(x) = L, x ? infinity, where L is in R. Then there exists an ?> 0 where given ?, there exist k(?) for all x > k then ? > max{1 , ([L]+1)/x} Therefore [xF(x) - L] ?. Therefore [F(x)] < ([L]+1)/x. Thus [F(x)-0] < ? Then lim F(x) =0 as x ? ?. This is what I have but it doesn't look right to me.

ID: 1891263 • Letter: P

Question

Prove that if lim (x=> c) f(x) exists, then f is bounded on some neighbourhood c. (The function f: D=> R (all real numbers) is bounded on some neighbourhood of c if there exists a sigma > 0 and M > 0 such that 0 < l x-c l < sigma, x E D, implies l f(x) l < M. Show complete solution and proof.

Explanation / Answer

Let F:(a,?) ?R is such that lim xF(x) = L, x ? infinity, where L is in R. Then there exists an ?> 0 where given ?, there exist k(?) for all x > k then ? > max{1 , ([L]+1)/x} Therefore [xF(x) - L] < 1 whenever x > ?. Therefore [F(x)] < ([L]+1)/x. Thus [F(x)-0] < ? Then lim F(x) =0 as x ? ?. This is what I have but it doesn't look right to me.

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Prove that if lim (x=> c) f(x) exists, then f is bounded on some neighbourhood c

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