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July 13th, 2017, 07:07 PM   #1
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finding limit((exp(-sin(x))-1)/(x),x,0) without L'Hôpital's rule?

Hi Guys

Can you help me with this problem?

limit((exp(-sin(x))-1)/(x),x,0)

I want to know how to solve this without L'Hospital's rule..

thank you )
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July 13th, 2017, 07:55 PM   #2
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A pair of MacLaurin series (for $\sin{(x)}$ and $e^x$) would be the obvious approach.

You could say that for small $x$ we have $\sin x \approx x$ but it's not so rigorous.
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July 13th, 2017, 08:05 PM   #3
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Thank you for getting back to me. This question was assigned as part of first calculus chp3. so I haven't learned what you talked about yet. I think there must be some trick to doing it which I cannot figure out.. We haven't even studied L'Hospital's Rule at this point!
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July 13th, 2017, 08:08 PM   #4
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I guess I can do it with linearization of the functions sin(x) and then exp(x) near 0 as you pointed it out..

Thank you again!
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July 13th, 2017, 08:33 PM   #5
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Without series or l'Hôpital, I guess you'll need to know
$$\lim_{x \to 0} \frac{\sin x}{x} = 1$$
and that \begin{align*}
\left. \frac{\mathrm d}{\mathrm d x}e^{-x} \right|_{x=0} &= \lim_{h \to 0} \frac{e^{-h} - e^0}{h} \\
\left. -e^{-x} \right|_{x = 0} = -1 &= \lim_{h \to 0} \frac{e^{-h} - 1}{h}
\end{align*}

With that, I suppose we say
\begin{align*}
\lim_{x \to 0} \frac{e^{-\sin{(x)}} - 1}{x} &= \lim_{x \to 0} \frac{e^{-\frac{\sin{(x)}}{x} \cdot x} - 1}{x} \\
&= \lim_{x \to 0} \frac{\left(e^{\frac{\sin{(x)}}{x}}\right)^{-x} - 1}{x}
\end{align*}
Then as $x \to 0$ $e^{\frac{\sin{(x)}}{x}} \to e^1 = e$ and thus the expression tends to $-1$.

It's not completely rigorous, but I guess it's close enough.
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Last edited by v8archie; July 13th, 2017 at 08:53 PM.
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July 18th, 2017, 04:24 AM   #6
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thank you very much for showing this approach to me. It was very interesting how you used the definition of derivative to find the limit! Awesome )
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July 18th, 2017, 04:56 AM   #7
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Different, but equivalent to v8archie's:

$$\lim_{x\rightarrow 0} \frac{e^{-\sin(x)} - 1}{x} = \lim_{x\rightarrow 0} \frac{e^{-\sin(x)} - 1}{\sin(x)} \frac{\sin(x)}{x} = \lim_{\theta\rightarrow 0}\frac{e^{-\theta} - 1}{\theta}\lim_{x\rightarrow 0} \frac{\sin(x)}{x}$$

Then this is easily seen to be $-1$.
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July 18th, 2017, 05:36 PM   #8
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I can see the change of variable that you've done! it's very nice, thank you!
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