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 February 7th, 2018, 09:58 AM #1 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond What do you get if you apply L'Hôpital's rule? Thanks from Jaket1 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: 3,264 Thanks: 902 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 $\displaystyle 1+ x/2+ \cdot\cdot\cdot+ x^{n-1}/n!+ \cdot\cdot\cdot$. The limit of that, as x goes to 0, should be obvious. Thanks from Jaket1 Last edited by Country Boy; February 7th, 2018 at 02:27 PM. February 8th, 2018, 05:02 AM #4 Senior Member   Joined: Jan 2016 From: Blackpool Posts: 104 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: 7,674 Thanks: 2654 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. Thanks from Jaket1 February 8th, 2018, 03:01 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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: 7,958 Thanks: 1146 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
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Quote:
 Originally Posted by Country Boy L'Hopital's rule certainly is a "standard formula using derivatives": If f(a)= 0 and g(a)= 0 then $\lim_{x\to a} \frac{f(x)}{g(x)}= \lim_{x\to a} \frac{f'(x)}{g'(x)}$.
It certainly isn't. In fact, applying L'hôpital's rule to this problem is completely circular since its application is predicated upon already knowing the very limit you want to compute. This is a common abuse of L'hôpital's rule. You can not use it to compute derivatives without committing a logical sin.

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: 7,958 Thanks: 1146 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
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Quote:
 Originally Posted by greg1313 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!
Are you kidding here? I'm having trouble believing this isn't a joke. In case it isn't, you should remind yourself what hypothesis is needed to use L'hôpital's rule. If that still doesn't do it, consider what you "learn" by computing a derivative with L'hôpital's rule.

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. Tags derivative, limit, question 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 daigo Calculus 1 May 17th, 2014 08:06 PM Quaid Calculus 10 September 16th, 2012 11:00 AM kaka2012sea Real Analysis 1 November 11th, 2011 08:52 AM kaka2012sea Calculus 1 November 11th, 2011 02:52 AM steezysam Calculus 3 November 4th, 2011 03:03 PM

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