March 11th, 2017, 12:39 PM  #1 
Newbie Joined: Mar 2017 From: San Ramon Posts: 1 Thanks: 0  Please suggest solution for limit of (e^x+e^x2)/x^2 as x goes to zero
Hi All, Can you please suggest how can I solve this : Finding the limit of (e^x+e^x2)/x^2 as x goes to zero 
March 11th, 2017, 01:19 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,783 Thanks: 1025 Math Focus: Elementary mathematics and beyond 
Use L'Hôpital's rule.
Last edited by skipjack; March 11th, 2017 at 03:39 PM. 
March 11th, 2017, 01:25 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,494 Thanks: 578 
You need to do it twice.

March 11th, 2017, 03:13 PM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra 
This problem is perfect for Maclaurin series. Far easier than l'Hôpital too.

March 11th, 2017, 05:14 PM  #5 
Member Joined: Jan 2016 From: Athens, OH Posts: 86 Thanks: 44 
Archie, I understand you have a personal vendetta against l'hôpital, but for beginners, I think this is much easier to understand than a series solution. Besides, it's much easier to type. Last edited by skipjack; March 11th, 2017 at 07:28 PM. 
March 11th, 2017, 06:10 PM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,783 Thanks: 1025 Math Focus: Elementary mathematics and beyond 
The limit is one.

March 11th, 2017, 06:17 PM  #7 
Member Joined: Jan 2016 From: Athens, OH Posts: 86 Thanks: 44 
In terms of age, I'm definitely not a beginning student. However, I managed to come up with the wrong answer in two different ways. So here's the correction: 
March 11th, 2017, 07:18 PM  #8  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,268 Thanks: 2434 Math Focus: Mainly analysis and algebra  Quote:
It might be easier to type if you pick the worst possible way to write the series! If you write it sensibly, it is very easy and extremely clear. Unless you believe that addition and subtraction is more difficult than differentiation. $$\begin{align*}e^x &= 1 + x +\tfrac12x^2 + \mathcal O(x^3) \\ \implies e^{x} &= 1  x +\tfrac12x^2 + \mathcal O(x^3) \\[8pt] \frac{e^x + e^{x} 2}{x^2} &= \frac{(1 + x +\tfrac12x^2) + (1  x +\tfrac12x^2) 2}{x^2} \\ &= \frac{x^2 + \mathcal O(x^3) }{x^2} \\ &= 1 + \mathcal O(x) \\[12pt] \lim_{x \to 0} \frac{e^x + e^{x} 2}{x^2} &= \lim_{x \to 0} \big( 1 + \mathcal O(x) \big) = 1 \end{align*}$$ Last edited by skipjack; March 11th, 2017 at 07:25 PM.  
March 11th, 2017, 07:26 PM  #9 
Global Moderator Joined: Dec 2006 Posts: 18,841 Thanks: 1564 
How do you know the power series for $e^x$?

March 12th, 2017, 07:11 PM  #10 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,783 Thanks: 1025 Math Focus: Elementary mathematics and beyond 
$$\lim_{x\to0}\frac{e^x+e^{x}2}{x^2}=2\lim_{x\to0}\frac{\cosh x1}{x^2}$$ Now use the Taylor series for $\cosh x$, $\sum_{n=0}^\infty\frac{x^{2n}}{(2n)!}$ 

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