February 7th, 2018, 09:58 AM  #1 
Member Joined: Jan 2016 From: Blackpool Posts: 95 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 05:02 PM. 
February 7th, 2018, 10:28 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,807 Thanks: 1045 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 05:02 PM. 
February 7th, 2018, 02:22 PM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
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 02:27 PM. 
February 8th, 2018, 05:02 AM  #4 
Member Joined: Jan 2016 From: Blackpool Posts: 95 Thanks: 2 
Is l'hôpital's rule classed as a "standard formula for derivatives"? Thanks.
Last edited by skipjack; February 8th, 2018 at 05:03 PM. 
February 8th, 2018, 07:24 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,308 Thanks: 2443 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, 03:01 PM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,165 Thanks: 867 
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 05:04 PM. 
February 8th, 2018, 04:19 PM  #7 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,807 Thanks: 1045 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 05:04 PM. 
February 8th, 2018, 04:25 PM  #8  
Senior Member Joined: Sep 2016 From: USA Posts: 379 Thanks: 205 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 05:05 PM.  
February 8th, 2018, 04:31 PM  #9 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,807 Thanks: 1045 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 05:06 PM. 
February 8th, 2018, 05:06 PM  #10  
Senior Member Joined: Sep 2016 From: USA Posts: 379 Thanks: 205 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 05:08 PM.  

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