My Math Forum finding limit((exp(-sin(x))-1)/(x),x,0) without L'Hôpital's rule?

 Calculus Calculus Math Forum

 July 13th, 2017, 06:07 PM #1 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 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 )
 July 13th, 2017, 06:55 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra 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. Thanks from mgho
 July 13th, 2017, 07:05 PM #3 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 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!
 July 13th, 2017, 07:08 PM #4 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 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!
 July 13th, 2017, 07:33 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,330 Thanks: 2457 Math Focus: Mainly analysis and algebra 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. Thanks from mgho Last edited by v8archie; July 13th, 2017 at 07:53 PM.
 July 18th, 2017, 03:24 AM #6 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 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 )
 July 18th, 2017, 03:56 AM #7 Senior Member   Joined: Oct 2009 Posts: 428 Thanks: 144 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 $-1$. Thanks from mgho
 July 18th, 2017, 04:36 PM #8 Newbie   Joined: Jun 2017 From: Long Beach Posts: 20 Thanks: 0 I can see the change of variable that you've done! it's very nice, thank you!

### l'Hôpital

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post mgho Trigonometry 5 June 28th, 2017 08:31 AM spidermanizblu3 Calculus 2 November 14th, 2016 01:15 PM idontknow Pre-Calculus 12 August 22nd, 2016 06:48 PM efking Calculus 2 October 31st, 2015 07:08 PM MathsIcy Calculus 5 February 2nd, 2015 06:17 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top