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

Sqrt of square is absolute value

For each n ≥ 1 define fn:R→Rf_{n} : \mathbb{R} \to \mathbb{R} by fn(x)=x2/x2+1/n,x∈Rf_{n}(x) = x^{2} / \sqrt{x^{2} + 1/n}, x \in \mathbb{R}, where √\surd denotes the non-negative square root. Wherever limn→∞fn(x)lim_{n}\to\infty f_{n}(x) exists, denote it by f(x). Which of the following statements is true?

  1. A.There exists x∈Rx \in \mathbb{R} such that f(x) is not defined
  2. B.f(x) = 0 for all x∈Rx \in \mathbb{R}
  3. C.f(x) = x for all x∈Rx \in \mathbb{R}
  4. D.f(x) = |x| for all x∈Rx \in \mathbb{R}✓ Correct

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, and why this one tests standard counterexample.

See pricing

50 are analysed free — try those first.

The trap it tests

Standard counterexample

There is a canonical object that settles this. Recognising it is the whole question.

Drill statements like this

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

More on this topic

The chapter behind this: Uniform convergence — what it buys and how to test it — free to read

From Sequences and Series of Functions › Pointwise vs uniform convergence, M-test, Dini

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