February 7th, 2018, 10:58 AM  #1 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2  derivative limit question:
Use standard formulae for derivatives to calculate the limit of \[\frac{e^{z}1}{z}\] as z tends to 0 The problem I am having with this question is how I am supposed to implement standard formulae for derivatives to help approach this question. I know that the limit as z tends to 0 for this function is undefined as you cannot divide by 0. Thanks. Last edited by skipjack; February 8th, 2018 at 06:02 PM. 
February 7th, 2018, 11:28 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
What do you get if you apply L'Hôpital's rule?
Last edited by skipjack; February 8th, 2018 at 06:02 PM. 
February 7th, 2018, 03:22 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
Another way to do this is to use the definition of "the derivative of f(x) at x= 0": $\displaystyle \lim_{h\to 0} \frac{f(h) f(0)}{h}$. With $\displaystyle f(x)= e^x$ that is $\displaystyle \lim_{h\to 0}\frac{e^h 1}{h}$. And what is the derivative of $\displaystyle e^x$? At x= 0? Yet another way: The MacLaurin series for $\displaystyle e^x$ is $\displaystyle 1+ x+ x^2/2+ \cdot\cdot\cdot+ x^n/n!+ \cdot\cdot\cdot$ so the MacLaurin series for $\displaystyle e^x 1$ is $\displaystyle x+ x^2/2+ \cdot\cdot\cdot+ x^n/n!$. Divide that by x to get $\displaystyle 1+ x/2+ \cdot\cdot\cdot+ x^{n1}/n!+ \cdot\cdot\cdot$. The limit of that, as x goes to 0, should be obvious. Last edited by Country Boy; February 7th, 2018 at 03:27 PM. 
February 8th, 2018, 06:02 AM  #4 
Senior Member Joined: Jan 2016 From: Blackpool Posts: 100 Thanks: 2 
Is l'hôpital's rule classed as a "standard formula for derivatives"? Thanks.
Last edited by skipjack; February 8th, 2018 at 06:03 PM. 
February 8th, 2018, 08:24 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
No, this is the standard formula for derivatives: $\displaystyle f'(x) = \lim_{h\to 0} \frac{f(x+h) f(x)}{h}$ You want to look at Country Boy's post.

February 8th, 2018, 04:01 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
L'Hôpital's rule certainly is a "standard formula using derivatives": If f(a)= 0 and g(a)= 0 then $\lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a} \frac{f'(x)}{g'(x)}$. Last edited by skipjack; February 8th, 2018 at 06:04 PM. 
February 8th, 2018, 05:19 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
Jaket1 stated that the limit is "undefined". L'Hôpital's rule is an easy out.
Last edited by skipjack; February 8th, 2018 at 06:04 PM. 
February 8th, 2018, 05:25 PM  #8  
Senior Member Joined: Sep 2016 From: USA Posts: 499 Thanks: 277 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
As a general rule, anytime the denominator is a linear function, you are taking a derivative of something, so just figure out what derivative it is to evaluate the limit. Last edited by skipjack; February 8th, 2018 at 06:05 PM.  
February 8th, 2018, 05:31 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,881 Thanks: 1088 Math Focus: Elementary mathematics and beyond 
I don't think so. How is the derivation of L'Hôpital's rule related to any specific derivative? If anything, using the definition of the derivative is redundant, as the limit we arrive at through doing so is exactly the limit we wish to compute! Last edited by skipjack; February 8th, 2018 at 06:06 PM. 
February 8th, 2018, 06:06 PM  #10  
Senior Member Joined: Sep 2016 From: USA Posts: 499 Thanks: 277 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
Suppose $f \in C^1$, then applying L'hôpital's rule to the derivative we have \[ \lim_{h \to 0} \frac{f(x+h)  f(x)}{h} = \lim_{h \to 0} \frac{\frac{d}{dh}f(x+h)  f(x)}{\frac{d}{dh}h} = \lim_{h \to 0} \frac{f'(x+h)}{1} = f'(x) \] which only tells us we were trying to compute a derivative. In other words, L'hôpital's rule can't compute derivatives. Claiming the derivative is "redundant" is putting the cart before the horse. Last edited by skipjack; February 8th, 2018 at 06:08 PM.  

Tags 
derivative, limit, question 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Question about limit definition of partial derivative  daigo  Calculus  1  May 17th, 2014 09:06 PM 
Limit of a Derivative Question  Quaid  Calculus  10  September 16th, 2012 12:00 PM 
derivative/limit  kaka2012sea  Real Analysis  1  November 11th, 2011 09:52 AM 
need help derivative/limit  kaka2012sea  Calculus  1  November 11th, 2011 03:52 AM 
Help with a derivative/limit question  steezysam  Calculus  3  November 4th, 2011 04:03 PM 