Solve for f: lim y → x ( x − y ) 2 f ( x ) − f ( y ) = 1

Question

Solve for f:      lim          y      →      x                  (      x      −      y              )        2                    f      (      x      )      −      f      (      y      )        =  1

Ronald Hickman · Accepted Answer

There is a mistake, which lies in jumping from

lim_{y \to x} \frac{(y - x)^{2}}{f (y) - f (x)} = 1 to lim_{y \to x} (y - x)^{2} = lim_{y \to x} f (y) - f (x) .

For instance,

lim_{x \to 0} \frac{\sin (\frac{1}{x})}{\sin (\frac{1}{x})} = 1,

but the limit

lim_{x \to 0} \sin (\frac{1}{x})

does not exist.
Note that

\begin{aligned} lim_{y \to x} \frac{(y - x)^{2}}{f (y) - f (x)} = 1 & ⟺ lim_{y \to x} \frac{f (y) - f (x)}{(y - x)^{2}} = 1 \\ ⟺ lim_{y \to x} \frac{f (y) - f (x)}{y - x} \cdot \frac{1}{y - x} = 1. \end{aligned}

But, since

lim_{y \to x} | \frac{1}{y - x} | = \infty

, it follows from this that

lim_{y \to x} \frac{f (y) - f (x)}{y - x} = 0

. In other words,

f^{'}

is the null function. And so, yes, f must be constant. But note that I only proved that if such a function exits, then it must be constant. Since no constant function is a solution of your problem, the problem has no solution.

Solve for f: <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-TeXAto

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