1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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... $\endgroup$ – Did May 17 '17 at 15:52
  • 5
    $\begingroup$ If you want to find the Taylor series why don't you just divide the Taylor series for $\sin x$ by $x$? $\endgroup$ – Zain Patel May 17 '17 at 15:53
  • $\begingroup$ Are you asking for the second derivative or for the third derivative? $\endgroup$ – Did May 17 '17 at 16:22
  • $\begingroup$ The second derivative is enough. Doesn't $\frac{\cos x}{x^2}$ as $x \to 0$ diverge? $\endgroup$ – user30523 May 17 '17 at 21:54
  • $\begingroup$ Yes -- and this is irrelevant. $\endgroup$ – Did May 19 '17 at 20:05
1
$\begingroup$

HINT: A better method is to note that function is even( check domain). what does that imply? I am sure you can take it from there. Keep note of differentiability of function using first principle, first

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.