# Why is the given transformation bijective and bicontinuous?

The following is an explanation that I've found on this site for finding the derivative of $$f(x) = x^2 \sin(1/x)$$ using the derivative definition. I was hoping that someone could clarify exactly why we set $$k = \frac{-h}{2x(x+h)}$$ and then solve for $$h$$ to come to the conclusion that the transformation is bijective and bicontinuous. Why exactly is that important? And if it is bijective and bicontinuous, how would one arrive at that conclusion?

\begin{align} f(x+h)-f(x)&=(x+h)^2\sin\frac{1}{x+h}-x^2\sin\frac{1}{x}\\ &=x^2\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2hx\sin\frac{1}{x+h}+h^2\sin\frac{1}{x+h} \end{align} When you divide by $$h$$ you get $$\frac{f(x+h)-f(x)}{h}= \frac{x^2}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)+2x\sin\frac{1}{x+h}+h\sin\frac{1}{x+h}$$ and you're left with computing $$>\lim_{h\to0}\frac{1}{h}\left(\sin\frac{1}{x+h}-\sin\frac{1}{x}\right)$$ because the second summand above tends to $$2x\sin(1/x)$$ and the third summand tends to $$0$$. You'll reinsert $$x^2$$ later.

Now you can use the identity $$\sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$$ Setting $$\alpha=1/(x+h)$$ and $$\beta=1/x$$, we have $$\frac{\alpha+\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}+\frac{1}{x}\right)= \frac{2x+h}{2x(x+h)}$$ while $$\frac{\alpha-\beta}{2}=\frac{1}{2}\left(\frac{1}{x+h}-\frac{1}{x}\right)= \frac{-h}{2x(x+h)}$$ so your limit is $$\lim_{h\to0}\frac{2}{h}\cos\frac{2x+h}{2x(x+h)}\sin\frac{-h}{2x(x+h)}$$ But $$\lim_{h\to0}2\cos\frac{2x+h}{2x(x+h)}=2\cos\frac{1}{x}$$ so we just need to compute $$\lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)}$$ Set $$k=-h/(2x(x+h))$$, so $$2x^2k+2xhk=-h$$ or $$h(2xk+1)=-2x^2k$$ that lends $$h=-\frac{2x^2k}{2xk+1}$$ So this transformation is bijective (and bicontinuous) and $$h\to0$$ implies $$k\to0$$ so your limit is $$\lim_{h\to0}\frac{1}{h}\sin\frac{-h}{2x(x+h)}= \lim_{k\to0}-\frac{2xk+1}{2x^2k}\sin k= \lim_{k\to0}-\frac{2xk+1}{2x^2}\frac{\sin k}{k}=-\frac{1}{2x^2}$$ In the end we get $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}= x^2\left(-\frac{1}{2x^2}\right)\cos\frac{1}{x}+2x\sin\frac{1}{x}= -\cos\frac{1}{x}+2x\sin\frac{1}{x}$$

This assumes, of course, $$x\ne0$$. If your function is defined by $$f(x)=\begin{cases} x^2\sin\dfrac{1}{x}&\text{for x\ne0}\\[1ex] 0&\text{for x=0} \end{cases}$$ we are left with the derivative at $$0$$, that is $$\lim_{h\to0}\frac{f(h)-f(0)}{h}= \lim_{h\to0}\frac{1}{h}h^2\sin\frac{1}{h}= \lim_{h\to0}h\sin\frac{1}{h}=0$$ with an easy application of the squeeze theorem.