 My Math Forum Need help with proving conversion of a limit
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 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^t-t-1 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 two-sided 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} Tags conversion, limit, proving Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post FreaKariDunk Real Analysis 3 November 19th, 2012 07:43 PM wawar05 Calculus 1 November 1st, 2012 04:07 AM moses Real Analysis 1 March 27th, 2011 09:57 AM mfetch22 Calculus 3 August 25th, 2010 07:41 AM oddlogic Calculus 2 January 25th, 2010 02:04 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.       