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May 12th, 2016, 05:39 AM  #1 
Newbie Joined: Jul 2013 Posts: 1 Thanks: 0  Need help with proving conversion of a limit
Prove that e^tt1 lim  = 1/2 t>0 t^2 without using L'Hospital's theorem. Help! I tried anything i can think of! 
May 12th, 2016, 05:46 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
So this is $\displaystyle \lim_{x\to 0}\frac{e^t t 1}{t^2}$? Obviously the simplest thing to do is to use L'Hopital's rule but if you don't want to, replace $\displaystyle e^t$ with its expression as a MacLaurin series: $\displaystyle \lim_{x\to 0}\frac{1+ t+ t^2/2+ t^3/6+ \cdot\cdot\cdot t 1}{t^2}= \lim_{t\to 0}\frac{t^2/2+ t^3/6+ \cdot\cdot\cdot}{t^2}= \lim_{t\to 0}\frac{1}{2}+ \frac{t}{6}+ \cdot\cdot\cdot$.

May 12th, 2016, 06:56 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
How is differentiating simpler than replacing with the series expansion and cancelling terms?

May 12th, 2016, 08:48 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
I now have a solution without series.

May 12th, 2016, 09:46 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
Here we go. Suppose that $\lim \limits_{t \to 0} {\mathrm e^t  t  1 \over t^2} = L$. Then we write $t=u$ to get $\lim \limits_{u \to 0} {\mathrm e^{u} +u  1 \over u^2} = L$, which is valid because we have a twosided limit. We can write both of these in terms of different variables: $x$ say; and sum them. $$\begin{aligned}2L &= \lim_{x \to 0} {\mathrm e^x  x  1 \over x^2} + \lim \limits_{t \to 0} {\mathrm e^{x} +x  1 \over x^2} & &= \lim_{x \to 0} \left({\mathrm e^x  x  1 \over x^2} + {\mathrm e^{x} +x  1 \over x^2}\right) \\ &= \lim_{x \to 0} {\mathrm e^x 2 +\mathrm e^{x} \over x^2} & &= \lim_{x \to 0} {\left(\mathrm e^{\frac12x}  \mathrm e^{\frac12x}\right)^2 \over x^2} \\ &= \left( \lim_{x \to 0} {\mathrm e^{\frac12x}  \mathrm e^{\frac12x} \over x}\right)^2 & &= \left( \lim_{x \to 0} {\left(\mathrm e^{\frac12x}1\right)  \left(\mathrm e^{\frac12x}1\right) \over x}\right)^2 \\ &= \left( \lim_{x \to 0} {\mathrm e^{\frac12x}1 \over x}  \lim_{x \to 0} {\mathrm e^{\frac12x}1 \over x}\right)^2 & &= \left( \left. {\mathrm d \over \mathrm d x} \mathrm e^{\frac12x}\right_{x=0}  \left. {\mathrm d \over \mathrm d x} \mathrm e^{\frac12x}\right_{x=0} \right)^2 \\ &= \left(\left.\frac12\mathrm e^{\frac12x}\right_{x=0} + \left.\frac12\mathrm e^{\frac12x}\right_{x=0}\right)^2 & &= \left(\frac12 + \frac12\right)^2 \\ &= 1 \\ L &= \frac12 \end{aligned}$$ 

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conversion, limit, proving 
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