Let f : D→R be a function and let c be a limit point of D.?

Assume f(x)→L as x→c：


(a) Prove that |f(x)| → |L| as x → c


(b) If f(x)%26gt;= 0 for all x ∈ D, prove that square root of f(x) →square root of L as x → c

Let f : D→R be a function and let c be a limit point of D.?
I'll give this proof under the assumption that D is supposed to be a subset of R. If D is instead a general topological space, then replace all references to 0%26lt;|x-c|%26lt;δ with x∈E, where E is a suitable punctured neighborhood of c (which neighborhood will be obvious from the context).





a: The key to this proof is to note that for all real numbers s and t, ||s| - |t|| ≤ |s-t|. This is easily derived from the triangle inequality thus: note that |s| = |s-t + t| ≤ |s-t| + |t|, so |s| - |t| ≤ |s-t|. By the same logic, |t| - |s| ≤ |t-s| = |s-t|. Now, ||s| - |t|| is either equal to |s| - |t| or |t| - |s|, and in either case is less than or equal to |s-t|.





With that proved, we proceed. Let ε%26gt;0 be arbitrary. Since f(x)→L, ∃δ%26gt;0 s.t. 0%26lt;|x-c|%26lt;δ ⇒ |f(x) - L| %26lt; ε. But then 0 %26lt; |x-c| %26lt; δ ⇒ ||f(x)| - |L|| %26lt; |f(x) - L| %26lt; ε, and since we can find such δ for any ε%26gt;0, it follows that |f(x)| → |L| as x→c.





b: The key to this is to prove that for any two nonnegative real numbers s and t, |√s - √t| ≤ √|s-t|. This is a less commonly used inequality than the first one, but not difficult to derive. We have:





|√s - √t| = |(s-t)/(√s + √t)| = |s-t|/(√s + √t)





Now, we consider two cases. If s≥t, then we have √|s-t| = √(s-t) ≤ √s ≤ √s + √t (since s and t are both nonnegative). If t≥s, we have √|s-t| = √(t-s) ≤ √t ≤ √t + √s = √s + √t. In either case, √|s-t| ≤ √s + √t, so 1/(√s + √t) ≤ 1/√|s-t|, and thus:





|s-t|/(√s + √t) ≤ |s-t|/√|s-t| = √|s-t|





Combining this with the first line yields |√s - √t| ≤ √|s-t|.





With that proved, let ε%26gt;0 be arbitrary. Then since f(x) → L, ∃δ%26gt;0 s.t. 0 %26lt; |x-c| %26lt; δ ⇒ |f(x) - L| %26lt; ε². But since f(x) and L are both nonnegative real numbers, we have by the inequality we derived that √f(x) - √L ≤ √|f(x) - L|, so 0 %26lt; |x-c| %26lt; δ ⇒ √f(x) - √L ≤ √|f(x) - L| %26lt; √(ε²) = ε. Since we can find such δ for any ε%26gt;0, it follows that √f(x) → √L as x→c.