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# Understand the problem

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.
##### Source of the problem
TIFR 2019 GS Part A, Problem 11
Analysis
Hard
##### Suggested Book
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 $$t \to 1$$.
1. So, for $$\epsilon >0,\exists \delta>0$$ s.t $$|g(t)|<\epsilon, \forall |x-1|<\delta$$
$$| \int_0^1 nx^ng(x)dx|\leq |\int_{0}^{1-\delta}nx^ng(x)dx|+|\int_{1-\delta}^{1} nx^ng(x)dx| \leq \int_{0}^{1-\delta}nx^n|g(x)|dx+\int_{1-\delta}^{1} nx^n|g(x)|dx =I_1+I_2$$. Now, $$I_1\to 0$$ as $$n \to \infty$$ as $$nx^n|g(x)|$$ is uniformly convergent to zero as $$n \to \infty$$ now $$I_2< \frac{n}{n+1}\epsilon$$ because of point 1) so we are done!!

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