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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 $e^x$.

Alexandros Voliotis
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6 Answers6

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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

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

user
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    (+1) This is the most efficient approach given the "standard limit" that was referenced. – Mark Viola Apr 18 '18 at 19:53
  • @MarkViola Thanks a lot Mark, I really appreciate your kind comment! Often the importance of the standard limits is highly underestimated. – user Apr 18 '18 at 19:57
  • Just curious as to how you suggest one prove that "standard limit." It can actually be shown directly from the limit definition of the exponential function. – Mark Viola Apr 18 '18 at 20:00
  • @MarkViola Yes of course that's fine or as an alternative, without an knowledge about derivative by $e^x-1=y\implies \frac{e^x-1}{x}=\frac{y}{log(1+y)}\to 1$ and $\frac{log(1+x)}{x}\to 1$ can be deived by the definition of $e$. – user Apr 18 '18 at 20:05
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The mean value theorem says

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

Since $c→0$, we get limit is $1$.

Takahiro Waki
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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 $0$.

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

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

  • You're right about being one L' Hospital away from the answer. But, I don't know taylor series yet. Also, by plugging into the $x$, 0 I got: $$e^0-(e^{\sin 0}\cos^20-(\sin0e^{\sin0}))-(\sin0e^{\sin0}+\cos0e^{\sin0})=$$ $$1-(11-(01))-1(11+11)=1-1-2=-2$$ – Alexandros Voliotis Apr 18 '18 at 19:35
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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 $1$.

zwim
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  • Good solution, but it is no different from the answer posted several minutes earlier by @gimusi ... You likely were typing you answer while that solution was posted and were unaware until thereafter. – Mark Viola Apr 18 '18 at 19:56
  • @MarkViola Yes, sometimes trying to explain with more words makes your post arriving behind people's oneliners rushing for credit, but it does not matter, I'm satisfied like this. – zwim Apr 18 '18 at 20:01
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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.

J.G.
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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 $$ \frac{e^{\sin x}-e^x}{x-\sin x} = \frac{\frac{x^3}{6}+o(x^3)}{\frac{x^3}{6}+o(x^3)} = \frac{\frac16+o(1)}{\frac16+o(1)} \rightarrow 1. $$

AHL
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  • In the last step it should be $$\frac{e^x-e^{\sin x}}{x-\sin x} = \frac{\frac{x^3}{6}+o(x^3)}{\frac{x^3}{6}+o(x^3)} = \frac{\frac16+o(1)}{\frac16+o(1)} \rightarrow 1$$ You don't deserve a downvote in my opinion even if Taylor's expansion is not necessary for the limit, you have properly derived the expansion. – user Apr 18 '18 at 22:30
  • Thank you, the minuses were clearly redundant. Edited. – AHL Apr 19 '18 at 16:30