Baffled with $\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$

Question

Calculate $$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}$$

Personal work:

$$\lim\limits_{x\to 0}{e^x-e^{\sin x} \over x-\sin x}=^{0 \over 0}\lim\limits_{x\to 0}{e^x-e^{\sin x}\cdot\cos x \over 1-\cos x}=^{0 \over 0}\lim\limits_{x\to 0}{{e^x-(e^{\sin x}\cdot\cos x-\sin x}\cdot e^{\sin x}\over \sin x})=\cdots$$

This gets to nowhere. Also, I substituted $t=e^{\sin x}$ but I could not replace the $e^x$.

Answer 1

Note that by standard limit for $t\to 0 \quad \frac{e^t-1}{t}\to 1$ since $(x-\sin x)\to 0\,$, we have

$${e^x-e^{\sin x} \over x-\sin x}=e^{\sin x}{e^{x-\sin x}-1 \over x-\sin x}\to 1\cdot 1=1$$

Answer 2

The numerator is approximately $e^x\,(1-e^{-x^3/6})\approx\frac{x^3}{6}e^x\approx\frac{x^3}{6}$ for small $x$. This approximation works for the denominator too, giving a limit of $1$. The cubic-over-cubic lowest-order expression also explains why using L'Hopital requires three iterations.

Answer 3

You are exactly one application of L'Hospital away from getting the answer, because the next denominator will be (non-vanishing!) $\cos x$. Alternatively, we can see that three applications of L'Hospital's rule are necessary because $x - \sin x$ vanishes to third order at $0$.

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Here's an approach using series with a detailed tracking of the error terms. We only need to keep terms up to the cubics, so we have $$\frac{1 + x + x^2 / 2 + x^3/6 + O(x^4) - \big(1 + \sin x + \sin^2 x / 2 + \sin^3 x / 6 + O(\sin^4 x)\big)}{x - \big(x - x^3/6 + O(x^5)\big)}$$

which is equal to

$$\frac{x + x^2/2 + x^3 / 6 + O(x^4) - (x - x^3/6 + O(x^5) + \frac 1 2(x^2 + O(x^4)) + x^3/6 + O(x^4)}{x^3 / 6 + O(x^5)}$$

which in turn simplifies as

$$\frac{x^3 / 6 + O(x^4)}{x^3/6 + O(x^5)} \to 1.$$

Answer 4

I propose a variant not using Taylor expansion.

Let set $u=x-\sin(x)\to 0$ when $x\to 0$

We have $$\dfrac{e^x-e^{\sin(x)}}{x-\sin(x)}=\dfrac{e^x-e^{x-u}}u=e^x\left(\dfrac{1-e^{-u}}u\right)$$

• the first term $e^x\to 1$
• the second term has limit $-f'(0)=1$ where $f(u)=e^{-u}$

So the overall limit is $1$.

Answer 5

The mean value theorem says

$\dfrac {e^x-e^{y}}{x-y}=e^c$

Since $c→0$, we get limit is $1$.

Answer 6

Use the third-order Maclaurin formulae $$ e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{6}+o(x^3), $$ $$ \sin x = x-\frac{x^3}{6}+o(x^3), $$ $$ e^{\sin x} = e^{x-\frac{x^3}{6}+o(x^3)} = 1+\left(x-\frac{x^3}{6}+o(x^3)\right)+\frac{1}{2} \left(x-\frac{x^3}{6}+o(x^3)\right)^2+ \frac16 \left(x-\frac{x^3}{6}+o(x^3)\right)^3 + o\left(\left( x-\frac{x^3}{6}+o(x^3) \right)^3 \right)= $$ $$ 1+x-\frac{x^3}{6} + \frac{x^2}{2}+\frac{x^3}{6} + o(x^3); $$ when expanding, we took into account only the summands up to the 3rd order, everything else gone into $o(x^3)$. Hence $$ \frac{e^{\sin x}-e^x}{x-\sin x} = \frac{\frac{x^3}{6}+o(x^3)}{\frac{x^3}{6}+o(x^3)} = \frac{\frac16+o(1)}{\frac16+o(1)} \rightarrow 1. $$