Comments by "MC116" (@angelmendez-rivera351) on "Find the Limit of sqrt(x) as x Approaches Infinity" video.

  1. Let f : [0, ∞) —> R, where R is the ordered field of real numbers, and [0, ∞) := {t in R : t >= 0}, and f(x) = sqrt(x) everywhere. We say lim f(x) (x —> ∞) = L for some real number L, if and only if for all real numbers ε > 0, there exists some U > 0, such that if x > U, then |f(x) – L| < ε. |f(x) – L| < ε is equivalent to L – ε < sqrt(x) < L + ε. Since sqrt(x) >= 0, it follows that max(0, L – ε) < sqrt(x) < L + ε. This is equivalent to x >= 0 and max(0, L – ε)^2 < x < (L + ε)^2. Hence, if max(0, L – ε)^2 =< U, then if x > U, x > max(0, L – ε)^2. However, x < (L + ε)^2 for some ε, L, and x > U. Therefore, lim sqrt(x) (x —> ∞) = L is false for every real number L, which means lim sqrt(x) (x —> ∞) does not exist. On the other hand, for all real numbers B > 0, there does exist a real number U > 0, such that if x > U, then sqrt(x) > B. This would be the case whenever U >= B^2. Therefore, one may say that as x —> ∞, sqrt(x) —> ∞. This is just special notation to say what I said above.
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