My book's process of calculating $\lim_{x\to0} \frac{e^x-e^{-x}-2x}{x-\sin x}$ and my process don't match. Why?

This question came in the Khulna University admission exam 2017-18

Q) What is the value of $$\lim_{x\to0} \frac{e^x-e^{-x}-2x}{x-\sin x}$$?

(a) 0

(b) 1

(c) 2

(d) 3

My attempt:

We can apply L'Hopital's rule here:

$$\lim_{x\to0} \frac{e^x-e^{-x}-2x}{x-\sin x}$$

$$=\lim_{x\to0} \frac{e^x+e^{-x}-2}{1-\cos x}\tag{1}$$

$$=\lim_{x\to0} \frac{e^x-e^{-x}}{\sin x}$$

$$=\lim_{x\to0} \frac{e^x+e^{-x}}{\cos x}$$

$$=\frac{1+1}{1}$$

$$=\frac{2}{1}=2$$

So, (c).

Third-party question bank's attempt:

$$\lim_{x\to0} \frac{e^x-e^{-x}-2x}{x-\sin x}$$

$$=\lim_{x\to0}\frac{2e^x-2}{1-\cos x}\tag{2}$$

$$=\lim_{x\to0}\frac{2e^x}{\sin x}$$

$$=\lim_{x\to0}\frac{2e^x}{\cos x}$$

$$=2$$

How did they write $$(2)$$? After $$(2)$$, they just used L'Hopital's rule, but I'm having trouble understanding how they arrived at $$(2)$$. The denominator in my $$(1)$$ and the denominator in $$(2)$$ match, so they probably used L'Hopital's to arrive at $$(2)$$ as well, but then why aren't the numerators matching in $$(1)$$ and $$(2)$$? Is it possible that they made a mistake in reaching $$(1)$$, and got the correct answer accidentally?

• I think your attempt is correct. $(2)$ is wrong, IMO.
– Feng
Commented Jul 14, 2022 at 5:42
• feedback: Even though providing context in Questions is important on this website, I feel that many of your questions (and their titles) benefit from being more distilled. It saves everyone's time, mostly yours. Doing so is also very good practice for writing in general (not just mathematical writing). Commented Jul 14, 2022 at 6:56

The third-party solution simply made a careless mistake at $$(2).$$
Notice that $$e^{x} - e^{-x} - 2x = x^{3}/3 + o(x^{4})$$ and $$x - \sin(x) = x^{3}/6 + o(x^{4})$$. Then we get: \begin{align*} \lim_{x\to 0}\frac{e^{x} - e^{-x} - 2x}{x - \sin(x)} = \lim_{x\to 0}\frac{x^{3}/3 + o(x^{4})}{x^{3}/6 + o(x^{4})} = \frac{1/3}{1/6} = 2 \end{align*}