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Evaluate $$\lim_{x \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x}$$

How do I find the limit without L'Hopital's rule?

It is well known that $$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$ So, if the limit were $$\lim_{(x - \sin x) \to 0} \frac{e^{x - \sin x} - 1}{x - \sin x}$$ the answer would be $1$. But here $x \to 0$, so I have no idea how to proceed.

Any help would be appreciated.

Thanks

Sourav Ghosh
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MangoPizza
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4 Answers4

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Simply note that $$x\to0\implies x-\sin x \to0$$

Cathedral
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Set $y=x-\sin x$.

For $x\to 0$ the new variable $y$ tends to $0$. So you can use the well-known limit theorem to conclude that

$$\lim_{x\to0} \frac{e^{x-\sin x}-1}{x-\sin x}=1$$

Sourav Ghosh
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InsideOut
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  • I don't understand this... $x - \sin x \neq x$. Sure $x - \sin x \to 0$ as $x \to 0$ but they aren't the same thing..? Is there a proof why we can substitute this? – MangoPizza Aug 06 '22 at 14:15
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    (Just so we don't have a misunderstanding I didn't downvote you. As soon as I saw a downvote, I actually upvoted.) – MangoPizza Aug 06 '22 at 14:20
  • You’re right, they are not the same thing. However they both approach to $0$ as $x\to0$ and this what really matters. Write $\lim_{y\to0}\frac{e^y-1}{y}=1$. In principle $y$ may depends on another variable, say $x$ and suppose $y(x)\to0$ as $x\to x_o$. Then you can write $\lim_{x\to x_o} \frac{e^{y(x)}-1}{y(x)}=1$. In your case, $y(x)=x-\sin x$ and $x_o=0$. – InsideOut Aug 06 '22 at 14:23
  • @MangoPizza Perhaps this thread will help: https://math.stackexchange.com/questions/1069642/finding-a-limit-using-change-of-variable-how-come-it-works – Cathedral Aug 06 '22 at 14:24
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    @Cathedral Thank you. This is exactly what I wanted. – MangoPizza Aug 06 '22 at 14:26
1

You already know that $$\lim_{x\to0}\frac{e^x-1}x=1$$

This is true not just for $x$ but for any function of $x$ if that is tending to zero i.e. $$\lim_{f(x)\to0}\frac{e^{f(x)}-1}{f(x)}=1$$

In general, if $f(x)$ is tending to zero when $x$ is tending to some 'a' then we say $$\lim_{x\to a}\frac{e^{f(x)}-1}{f(x)}=1$$

For example, $$\lim_{x\to2}\frac{e^{x-2}-1}{x-2}=1$$

aarbee
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Define $t=x-\sin(x)$, when $x\to0, ~t\to0$

$$\lim_{x\to0} \frac{e^{x-\sin x}-1}{x-\sin x}=\lim_{t\to0}\frac{e^t-1}{t}=\lim_{t\to0}\frac{e^t}{1}=1$$

MathFail
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