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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: 6,857 Thanks: 2230 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. 
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: 6,857 Thanks: 2230 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. 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 
Member Joined: Oct 2009 Posts: 97 Thanks: 33 
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 $1$. 
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!


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