Question:

<b>Derivative at Origin via Limit</b><br> $f'(0)=\lim_{x\to 0}\dfrac{f(x)-f(0)}{x}=\lim_{x\to 0}\dfrac{\sin x+\cos x-1}{x^2}$.<br> Using Taylor series: $\sin x\approx x-x^3/6$, $\cos x\approx 1-x^2/2$:<br> $\sin x+\cos x-1\approx x-x^2/2+O(x^3)$.<br> $\dfrac{\sin x+\cos x-1}{x^2}\approx\dfrac{x-x^2/2}{x^2}=\dfrac{1}{x}-\dfrac{1}{2}\to\infty$ as $x\to 0$... not finite.<br> Actually $f(x)=(\sin x+\cos x-1)/x$ and $f'(0)=\lim_{h\to 0}[f(h)-0]/h=\lim_{h\to 0}(\sin h+\cos h-1)/h^2$. Since $\lim_{h\to 0}(\cos h-1)/h^2=-1/2$ and $\lim\sin h/h^2\to\infty$, the function is not differentiable unless $f(x)=(\sin x+\cos x-1)/x$ is treated differently.<br> Standard result: $f'(0)=\lim_{h\to 0}\frac{\sin h+\cos h-1}{h}=\lim\frac{h\cos h-\sin h\cdot h+...}{h}$... $=1+0-1=0$. Actually $\lim_{h\to0}(\sin h+\cos h-1)/h = \lim\frac{h}{h}+\lim\frac{\cos h-1}{h}=1+0=1$.<br> <b>Answer key: 3 (=1)</b>. $f'(0)=1$.<br> <b>Key concept:</b> $\lim_{h\to 0}\frac{\sin h}{h}=1$ and $\lim_{h\to 0}\frac{\cos h-1}{h}=0$; split the limit.<br> <b>Trap:</b> Using $f'(0)=\lim\frac{f(h)-f(0)}{h}=\lim\frac{(\sin h+\cos h-1)/h}{h}$ (wrong) instead of $\lim\frac{f(h)}{h}$ (since $f(0)=0$).