Part BCSIR NET June 2024sqrt-of-square-is-absolute-value

Sqrt of square is absolute value

For each n ≥ 1 define $f_{n} : R \to R$ by $f_{n} (x) = x^{2} / x^{2} + 1/ n, x \in R$ , where $\sqrt$ denotes the non-negative square root. Wherever $l i m_{n} \to \infty f_{n} (x)$ exists, denote it by f(x). Which of the following statements is true?

A.There exists $x \in R$ such that f(x) is not defined
B.f(x) = 0 for all $x \in R$
C.f(x) = x for all $x \in R$
D.f(x) = |x| for all $x \in R$ ✓

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward.

See pricing

50 are analysed free — try those first.

Related counterexample: fₙ → f uniformly ⇒ fₙ′ → f′

Sqrt of square is absolute value

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