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Suppose $f: (0,1) \to \mathbb R$ differentiable such that $$\lim_{x\to0^+}f(x) = A$$ and $$\lim_{x\to0^+}xf'(x)=B.$$ Find the value of B.


My attempt:

We have $$(xf(x))' = f(x) + xf'(x) \ \ \ \ \ (1)$$ But $(xf(x))'$, in the limit $x\to0^+$, is $$\lim_{x\to0^+}\frac{xf(x)}{x} = A$$ So taking limits on both sides of $(1)$, we have $B = 0$.

Is my approach correct?

If it isn't correct, how to solve this?

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  • $\begingroup$ You are assuming that $xf'(x)$ is continuous at $0$. This is not given. $\endgroup$ Commented May 8, 2023 at 11:33
  • $\begingroup$ What justifies your claim $\lim_{x\to0^+}(xf(x))'=\lim_{x\to0^+}\frac{xf(x)}{x}?$ $\endgroup$ Commented May 8, 2023 at 11:35
  • $\begingroup$ @AnneBauval The definition of derivative $\endgroup$ Commented May 8, 2023 at 11:44
  • $\begingroup$ @geetha290krm but I just know the continuity at right hand side, and I only used this $\endgroup$ Commented May 8, 2023 at 11:45
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    $\begingroup$ @KennyWong OP's solution is nowhere near being correct. $\endgroup$ Commented May 8, 2023 at 11:49

3 Answers 3

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Note that the limit $\lim_{x\rightarrow 0^+}xf'(x)$ may not exist. For a counterexample, consider the function $f(x) = x\sin(1/x)$. However, if we are given that the limit exists then we may use L'Hopital's rule: $$A = \lim_{x\rightarrow 0^+}\frac{xf(x)}{x} = \lim_{x\rightarrow 0^{+}}\frac{xf'(x)+f(x)}{1} = A+B\implies B = 0.$$

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Let us prove by contradiction that $$B=0.$$ Assume wlog $B>0,$ and let $\epsilon:=\frac B2.$ For every positive $x$ sufficiently close to $0,$ we have both $$f(x)-A<\epsilon$$and (by the MVT)$$\frac{f(x)-A}x=f'(c_x)>\frac{B-\epsilon}{c_x}>\frac{B-\epsilon}x=\frac\epsilon x.$$ Whence the contradiction $\epsilon<\epsilon.$

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Your approach is essentially correct. But we need to address some fine points. Let's define a funtion $g$ such that $g(0)=0$ and $g(x) =xf(x) $ if $x\in(0,1)$. Then $g$ is continuous on $[0,1)$ based on hypotheses in question.

Now right derivative of $g$ at $0$ is $\lim_{x\to 0^+}g(x)/x=A$. Further we have $g'(x) =xf'(x) +f(x) $ for all $x\in(0,1)$ and hence $g'(x) \to A+B$ when $x\to 0^+$. It is standard result that derivatives can't have jump discontinuity and hence the right limit of $g'(x) $ must match the right derivative of $g $ at $0$. Hence $A=A+B$ ie $B=0$.

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