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# And you thought limits were always easy

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.

Note by Parth Thakkar
2 years, 8 months ago

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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) · 2 years, 7 months ago

Why should it be true, that "if $$f(x)$$ is not differentiable at $$x = 0$$, then $$\dfrac{f(x)}{x}$$ will not be continuous at $$x = 0$$."? 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 $$f(x) = |x|$$. 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 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 $$0$$ 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! :) · 2 years, 7 months 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 .. !! · 2 years, 7 months ago

@Calvin Lin , could you help me out with this one? Thanks. · 2 years, 8 months ago

Other problems I found interesting:

· 2 years, 8 months ago