Evaluating $\lim_{x\to 0} \frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x}$. Why is this solution wrong?

Question

the solution given in book is $1$ and by L'Hospital approach i do get one so what's the fault in method

Question:

$$\lim_{x\to 0} \frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x}$$

Work: $$\begin{aligned} \lim_{x\to 0} \frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x} & = \lim_{x\to 0} \frac{ x(\frac{e^x-1}{x}) +1 - \frac{1}{x(\frac{e^x-1}{x}) + 1}-2\frac{\ln(1+x)}{x}\cdot x}{x^2\cdot \frac{\sin x}{x} }\\& = \lim_{x\to 0} \frac{x + 1- \frac{1}{x+1} - 2x}{x^2}\\& = \lim_{x\to0}\frac{1 - x- \frac{1}{x+1}}{x^2}\\& = \lim_{x\to0} \frac{1 - x^2 - 1}{(x+1)x^2}\\& = \lim_{x\to 0}\frac{-1}{x+1}\\& = -1\end{aligned}$$

Why is this method wrong?

Answer 1

$$\lim_{x\to0}\frac{e^x-e^{-x}-2\ln(1+x)}{x\sin{x}}$$ Substitute the quadratic Taylor approximation of each function $$\lim_{x\to0}\frac{\left(1+x+\frac{x^2}{2}\right)-\left(1-x+\frac{x^2}{2}\right)-2\left(x-\frac{x^2}{2}\right)}{x(x)}=\lim_{x\to0}\frac{x^2}{x^2}=1$$ Your method is wrong because the limit formulas that you used are linear approximations, which omit higher degree terms such as $x^2$

Answer 2

As noticed, we can't replace single part of the expression as you have done. Sometimes it works but the result is not guaranteed in general. Therefore this way to proceed should be avoided.

In some sense, the way you have used corresponds to a first order approximation, which should be performed as follows

• $e^x=1+x+O(x^2)$
• $\log(1+x)=x+O(x^2)$
• $\sin x=x+O(x^2)$

to obtain

$$\frac{ e^x - e^{-x}-2\ln(1+x)}{x \sin x}=\frac{1+x-\frac1{1+x+O(x^2)}-2x+O(x^2)}{x^2+O(x^3)}=\frac{-x^2+O(x^2)}{x^2+O(x^3)}=\frac{-1+O(1)}{1+O(x)}$$

and here we see the issue hided behind your result: first order expansion doesn't suffice to obtain the result, indeed the term $O(1)$ in the numerator represents something which is not necessarly equal to zero.

This result explains also the importance to use the remainder term (big-O or little-o) when we use Taylor expansion in order to avoid wrong evaluations.

To obtain the correct limit, as indicated in the first given answer, we need to proceed with a second order expansion.

As an alternative we can also proceed without series expansion, using the results indicated here, as follows

$$\frac{e^x-e^{-x}-2\ln(1+x)}{x\sin{x}}=\frac x{\sin x}\frac{e^x-1-x-e^{-x}+1-x-2\ln(1+x)+2x}{x^2}=$$

$$=\frac x{\sin x}\left(\frac{e^x-1-x}{x^2}-\frac{e^{-x}-1+x}{x^2}-2\frac{\ln(1+x)-x}{x^2}\right)\to 1 \cdot \left(\frac12-\frac12-2\cdot -\frac12\right)=1$$