Let \(f\)* *be a continuous function on \([0,1]\). Then the limit \(\lim_{n \to \infty} \int_0^1 nx^nf(x)dx\) is equal to

- \(f(0)\)
- \(f(1)\)
- \(sup_{x\in [0,1]}f(x)\)
- The limit does not exist.

TIFR 2019 GS Part A, Problem 11

Analysis

Hard

Real analysis, Bartle,Sherbert

Do you really need a hint? Try it first!

Consider \( g(t)=f(t)-f(1)\). Can you prove limit \(\lim_{n \to \infty}| \int_0^1 nx^ng(x)dx|=0\) ?

Try to bound \(g(t)\) in the neighbourhood of \(1\) then try to prove hint 1

\(| \int_0^1 nx^ng(x)dx|\leq |\int_{0}^{1-\delta}nx^ng(x)dx|+|\int_{1-\delta}^{1} nx^ng(x)dx|\)

Once you are done with hint 1, what can you say about option 2?

\(g(t) \to 0\) as \(t \to 1\).

- So, for \(\epsilon >0,\exists \delta>0\) s.t \(|g(t)|<\epsilon, \forall |x-1|<\delta\)

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