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?
Best Answer
The third-party solution simply made a careless mistake at $(2).$
Such mistakes are called sign errors, and are easy to make, especially in exam situations.