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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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#### College Mathematics Program

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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