# Different taylor expansion order to calculate a limit

I have this limit to calculate :

$$\lim_{x\rightarrow 0}\frac{x \cos (x) - \sin (x)}{e^{x} - e^{-x} - 2x}$$

I need to know if my solution were true, I explain

• I use the second order of taylor expansion of $$\cos(x) = 1 - \frac{x^2}{2}$$
• I use the first order of taylor expansion of $$\sin(x) = x$$

$$\exp(x) = 1$$

$$\exp(-x) = 1$$

So I get

$$\lim_{x\rightarrow 0}\frac{x \cos (x) - \sin (x)}{e^{x} - e^{-x} - 2x} = \lim_{x\rightarrow 0}\frac{ \frac{-x^3}{2}}{ - 2x} = 0$$

• I think the answer is $-1$. – Rohan Dec 4 '16 at 15:19
• the searched limit is equal to $$-1$$ – Dr. Sonnhard Graubner Dec 4 '16 at 15:19
• @Rohan how you get this result ? my approach is it false? – Lamloumi Afif Dec 4 '16 at 15:20
• I got this limit in an assignment a few months back. Hence, I vaguely remember. Luckily, @Dr.SonnhardGraubner has confirmed it. – Rohan Dec 4 '16 at 15:22

We have

$$x\cos(x)-\sin(x)=x(1-\frac{x^2}{2})-x+\frac{x^3}{6}+x^3\epsilon_1(x)$$

$$=x^3\left(\frac{-1}{3}+\epsilon_1(x)\right).$$

and

$$e^x-e^{-x}-2x=1+x+\frac{x^2}{2}+\frac{x^3}{6}-1+x-\frac{x^2}{2}+\frac{x^3}{6}+x^3\epsilon_2(x)-2x$$

$$=x^3\left(\frac{1}{3}+\epsilon_2(x)\right).$$

thus the limit is $-1$.

• I agree with your method, but I need to know exactly what is the error in mine? – Lamloumi Afif Dec 4 '16 at 15:40
• @LamloumiAfif In lines 7,8,9 .you had to expand untill third order. – hamam_Abdallah Dec 4 '16 at 15:41

Your error is that all Taylor's expansions have to be at the same order, otherwise your your results will be meaningless.

• Numerator: $\; x\cos x=x\Bigl(1-\dfrac{x^2}2+o(x^2)\Bigr)=x-\dfrac{x^3}2+o(x^3)$, hence $$x\cos x-\sin x=x-\dfrac{x^3}2+o(x^3)-\Bigl(x-\dfrac{x^3}6+o(x^3)\Bigr)=-\dfrac{x^3}3+o(x^3)\sim_0-\dfrac{x^3}3$$
• Denominator: $\;\mathrm e^x-\mathrm e^{-x}-2x=2(\sinh x-x)=2\Bigl(x+\dfrac{x^3}6+o(x^3)-x\Bigr)=\dfrac{x^3}3+o(x^3)\sim_0\dfrac{x^3}3$. Thus the fraction is equivalent near $0$ to $$\frac{-\dfrac{x^3}3}{\phantom{-}\dfrac{x^3}3}=-1.$$