Indeterminate Sums $\infty-\infty$

Question

Doing the taylor expansion around $x^*=0$ for $\frac{\sin(x)}{x}-1$ I have found $$f'(x)=\frac{\cos(x)}{x}-\frac{\sin(x)}{x^2}$$ $$f''(x)=-\frac{\sin(x)}{x}+\frac{2\sin(x)}{x^3}-\frac{2\cos(x)}{x^2}$$ How can I formally show that both have well defined limits from both sides in the point $x^*=0$ with $f'(x^*)=0$ and $f''(x^*)=-1/3$? (These limits are implied by graphing). My problem is that $\infty-\infty$ crops up and I don't know how to deal with that particular indeterminate form.

Answer 1

Note that$$f(x)=\frac{\sin x}x-1=-\frac{x^2}{3!}+\frac{x^4}{5!}-\cdots$$So,\begin{align}f'(x)&=-\frac13x+\frac1{30}x^4-\cdots\\f''(x)&=-\frac13+\frac2{15}x^3-\cdots\end{align}and so on… In particular,

• $\lim_{x\to0}f(x)=0$;
• $\lim_{x\to0}f'(x)=0$;
• $\lim_{x\to0}f''(x)=-\frac13$

and so on.

Answer 2

$$\frac{\sin x}{x}-1=1-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{840}....-1=-\frac{x^2}{6}+\frac{x^4}{120}-\frac{x^6}{840}+...$$

Answer 3

Your problem is to find first few terms of the Taylor series around $0$ for the function $$f(x) =\dfrac{\sin x} {x}, x\neq 0,f(0)=1$$ The simpler approach is to use Taylor series for $\sin x$ and then divide by $x$.

But if you want to go by definition then you need to evaluate the derivatives $f'(0),f''(0),f'''(0)$ and then get the desired terms of the Taylor series.

To do so you need to use definition of derivative. We have $$f' (0)=\lim_{x\to 0}\frac{f(x)-f(0)}{x}=\lim_{x\to 0}\frac{\sin x - x} {x^2} =0$$ Further $$f'(x) =\frac{x\cos x-\sin x} {x^2}, x\neq 0$$ and hence $$f''(0)=\lim_{x\to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x\to 0}\frac{x\cos x - \sin x} {x^3}$$ The above limit can be evaluated using LHospital Rule to get $f''(0)=-1/3$.