# $f'''(0)$ of $f(x)=\sin x/x,x \neq 0,1, x=0$

I want to find the Taylor polynomial at zero for $f(x)=\sin x/x \ \text{if} \ x \neq 0, 1 \ \text{if} \ x=0$ and struggle to find $f''(0).$ I can't just differentiate $\sin x/x$ because we are considering $0$.I have $f'(0)=\lim_{x \to 0} \frac{\sin x/x-1}{x}=0$ (l'Hopital) but how do I find the second derivative, I need a general expression for the derivative?

• If $x\ne0$, $$f'(x)=\frac{x\cos x-\sin x}{x^2}$$ hence $$f''(0)=\lim_{x\to0}\frac{x\cos x-\sin x}{x^3}$$ Can you compute this limit? Knowing that $$1-\cos x\sim\frac{x^2}2\qquad x-\sin x\sim\frac{x^2}6$$ suffices... – Did May 17 '17 at 15:52
• If you want to find the Taylor series why don't you just divide the Taylor series for $\sin x$ by $x$? – Zain Patel May 17 '17 at 15:53
• Are you asking for the second derivative or for the third derivative? – Did May 17 '17 at 16:22
• The second derivative is enough. Doesn't $\frac{\cos x}{x^2}$ as $x \to 0$ diverge? – user30523 May 17 '17 at 21:54
• Yes -- and this is irrelevant. – Did May 19 '17 at 20:05