And you thought limits were always easy

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

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

lim⁡r→0+(∫−1−rf(x)xdx+∫r1f(x)xdx) \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 f to show that the limit need not exist if it is not differentiable at 0 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.

#Calculus #Limits #Integration #Proofs #Integrals

Note by Parth Thakkar
7 years ago

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Comments

One 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 - 7 years ago

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Why should it be true, that "if f(x) f(x) is not differentiable at x=0 x = 0 , then f(x)x \dfrac{f(x)}{x} will not be continuous at x=0 x = 0 ."? I haven't found a counter example, but could you please give a reason?

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

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Now, the integrals ∫−1−rf(x)xdx \displaystyle \int_{-1}^{-r} \dfrac{ f(x) } {x } dx and ∫+r1f(x)xdx \displaystyle \int^{1}_{+r} \dfrac{ f(x) } {x } dx will equal −1+r -1 + r and 1−r 1 - r respectively. So, the limit is −1+1=0 -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 that there 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 0 will attain different values which makes the limit not to exist for f(x) 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 - 7 years ago

@Calvin Lin , could you help me out with this one? Thanks.

Parth Thakkar - 7 years ago

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 - 7 years ago
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