Another CMI problem: (and I could not solve this)

Consider a continuous function \( f : [-1,1] \rightarrow R \) which is differentiable at \( 0 \). Prove that the following limit exists:

\( \displaystyle \lim_{r \to 0^{+}} \left( \int_{-1}^{-r} \dfrac{ f(x) } {x } dx + \int_{r}^1 \dfrac{ f(x) } {x } dx \right) \)

Also give an example of a function \( f \) to show that the limit need not exist if it is not differentiable at \( 0 \).

In case there is a mistake in the question (I don't think there should be), please tell me. The problem is, I don't have the question paper and it hasn't yet been uploaded on their site. So if there is any error, it must have crept in due to my poor memory.

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TopNewestOne thing that i can interpret from this question is that if f(x) is not differentiable at x=0 then f(x)/x will not be continuous at x=0 i.e. the right hand side and left hand side of 0 will attain different values and which makes the limit not to exist for f(x) – Ramesh Goenka · 3 years, 1 month ago

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Also, regarding your second clain: Let's consider \( f(x) = |x| \). Clearly, this isn't differentiable at \( x = 0 \), and also, \( \dfrac{f(x)}{x} \) isn't continuous at \( x = 0 \). (Agrees with your first claim).

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Now, the integrals \( \displaystyle \int_{-1}^{-r} \dfrac{ f(x) } {x } dx \) and \( \displaystyle \int^{1}_{+r} \dfrac{ f(x) } {x } dx \) will equal \( -1 + r \) and \( 1 - r \) respectively. So, the limit is \( -1 + 1 = 0 \). Doesn't it exist?

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Obviously, this doesn't answer the question - which is,

given a differentiable function show that the limit exists, while my example shows thatthere exists a non-differentiable function such that the limit exists. But this does counter your second claim ( "the right hand side and left hand side of \( 0 \) will attain different values which makes the limit not to exist for \( f(x) \)"). I don't say that the right and and left hand sides are the same, but still, the limit does exist.If there's something wrong, I'd love to see it! :) – Parth Thakkar · 3 years, 1 month ago

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with that i think the limit will always exist since the integral can always be broken into two different integrals about a point where its not continuous ... !! since r tends to 0+ and if we want to prove that the limit doesn't exists, the expression in the numerator should be a polynomial of degree less than that of in the denominator .. !! – Ramesh Goenka · 3 years, 1 month ago

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@Calvin Lin , could you help me out with this one? Thanks. – Parth Thakkar · 3 years, 2 months ago

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Other problems I found interesting:

Polynomials? That sounds familiar

A fun problem - Find the formula of number of functions from a power set to another set

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