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let $f(x)= (x^3+2x)[\frac{x}{2}]$

[x]:floor function

then :

$$\lim\limits_{h\to 0}\frac{f(2-h^2)-f(2)}{h^2}=?$$

My try :

$$\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{h}=f'(2)$$

$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-f(2)}{x-2}=f'(2)$$

$$\lim\limits_{x\to 2}\frac{(x^3+2x)[\frac{x}{2}]-12}{x-2}=f'(2)$$

Now what ?

Mithlesh Upadhyay
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Almot1960
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    What is $[\frac x 2]$ ? The floor function ? – Claude Leibovici Mar 28 '17 at 13:50
  • Even when $f$ is differentiable, the identity $\lim\limits_{h\to 0}\frac{f(2-h)-f(2)}{h}=f'(2)$ is not true. At best, $-f'(2)$. That being said, you have no evidence that $f$ is differentiable at $x=2$ here. –  Mar 28 '17 at 13:53
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    By the way, your work is at a dead end now, because you decided to consider the non-existing limit as $x\to 2$ (or, in your notation, as $h\to 0$): the fact that $h^2\ge 0$ is crucial. –  Mar 28 '17 at 14:03
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    @G.Sassatelli It appears that the desired limit doesn't exist either. – Michael Burr Mar 28 '17 at 14:15
  • @MichaelBurr Indeed. In the high-school sense, it would be $+\infty$. –  Mar 28 '17 at 15:29
  • @G.Sassatelli Agreed (but it makes me cringe to describe the derivative of a jump discontinuity as an infinite limit ... without distributions). – Michael Burr Mar 28 '17 at 15:31

1 Answers1

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Observe that as $h\rightarrow 0$, $h^2\rightarrow 0^+$ since $h^2$ is always nonnegative. Therefore, we can write:

$$ \lim_{h\rightarrow 0}\frac{f(2-h^2)-f(2)}{h^2}=\lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k} $$ by setting $k=h^2$. Then, by substituting the formula for $f$, we get \begin{align*} \lim_{k\rightarrow 0^+}\frac{f(2-k)-f(2)}{k}&=\lim_{k\rightarrow 0^+}\frac{((2-k)^3+2(2-k))\left\lfloor\frac{2-k}{2}\right\rfloor-(2^3+4)\left\lfloor\frac{2}{2}\right\rfloor}{k}\\ &=\lim_{k\rightarrow 0^+}\frac{(-k^3+6k^2-14k+12)\left\lfloor\frac{2-k}{2}\right\rfloor-12}{k} \end{align*} Since, when $k$ is slightly larger than $0$, $2-k$ is slightly less than $2$, $\frac{2-k}{2}$ is slightly less than $1$, so its floor is $0$. This implies that we're looking at $$ \lim_{k\rightarrow0^+}\frac{-12}{k} $$ which does not exist.

On the other hand, $$ \lim_{h\rightarrow 0}\frac{f(2+h^2)-f(2)}{h^2} $$ does exist and the limit is $14$.

Michael Burr
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