My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum

Thanks Tree4Thanks
  • 2 Post By greg1313
  • 2 Post By Math Message Board tutor
LinkBack Thread Tools Display Modes
October 12th, 2014, 10:34 AM   #1
Joined: Oct 2014
From: Norway

Posts: 42
Thanks: 1

Can't figure out how to apply l'Hopital here.

How do you find lim as x->0 of 2/(e^(5x^-2) * (5x^3))? I tried l'Hôpital Rules, but when I try to derivate the e term, then I just get another instance of the problem... (one of the rules say that if lim as x->0 of f'(x)/g'(x) = L, then lim as x->0 of f(x)/g(x) = L).

I'm not restricted to l'Hôpital's rules though. Just thought they might help. Wolfram|Alpha says the answer is 0, but I want to understand the process.

Last edited by skipjack; November 1st, 2014 at 06:49 PM.
uint is offline  
October 12th, 2014, 10:50 AM   #2
Math Team
Joined: Dec 2013
From: Colombia

Posts: 7,690
Thanks: 2669

Math Focus: Mainly analysis and algebra
L'Hôpital doesn't apply because the expression is not an indeterminate form. It is 2/0.

If Wolfram Alpha says zero, you've made a mistake either here or there.
v8archie is offline  
October 12th, 2014, 11:06 AM   #3
Joined: Oct 2014
From: Norway

Posts: 42
Thanks: 1

Notice that 2/e^(5x^-2) = 2e^(-5/x^2).^3%29%29 I suppose you overlooked the negation of the exponent of x.
uint is offline  
October 12th, 2014, 12:19 PM   #4
Global Moderator
greg1313's Avatar
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,968
Thanks: 1152

Math Focus: Elementary mathematics and beyond
The limit may be written as

$\displaystyle \frac25\lim_{x\to0}\frac{x^{-3}}{e^{5x^{-2}}}$

L'Hôpital's rule:

$\displaystyle =\frac25\lim_{x\to0}\frac{-3x^{-4}}{-10x^{-3}e^{5x^{-2}}}$

$\displaystyle =\frac{3}{25}\lim_{x\to0}\frac{x^{-1}}{e^{5x^{-2}}}$

Differentiating again,

$\displaystyle =\frac{3}{250}\lim_{x\to0}\frac{x}{e^{5x^{-2}}}=0$
Thanks from topsquark and uint
greg1313 is offline  
October 12th, 2014, 01:29 PM   #5
Banned Camp
Joined: Jun 2014
From: Earth

Posts: 945
Thanks: 191

If you want to reduce the number of negative signs in the number of characters of the solution,
you could make a substitution in variables, such as:

$\displaystyle \lim_{x\to 0} \ \dfrac{2}{ \ e^{5x^{-2}}*5x^3 \ } \ $

Let $\displaystyle \ y \ = \ x^{-1} \ = \ \frac{1}{x}.$

And so $\displaystyle \ x \ = \ \frac{1}{y}.$

As $\displaystyle \ x\to 0, \ \ y\to \pm \ \infty.$

$\displaystyle \lim_{x\to 0} \ \dfrac{2}{ \ e^{5x^{-2}}*5x^3 \ } \ = $

$\displaystyle \lim_{y\to \ \pm \infty} \dfrac{2}{ \ e^{5y^2}*5\bigg(\frac{1}{y^3}\bigg)} \ \ = $

Move the fractional constant in front of the limit, and multiply the numerator and denominator by $\displaystyle \ y^3:$

$\displaystyle \bigg(\frac{2}{5}\bigg)\lim_{y\to \ \pm \infty} \ \dfrac{y^3}{ e^{5y^2} } \ = $

L'Hôpital's rule may be used now if you wish.
Thanks from topsquark and uint

Last edited by Math Message Board tutor; October 12th, 2014 at 01:34 PM.
Math Message Board tutor is offline  

  My Math Forum > College Math Forum > Calculus

apply, figure, lhopital

Thread Tools
Display Modes

Similar Threads
Thread Thread Starter Forum Replies Last Post
Can I apply L'hopital here? OriaG Calculus 1 February 8th, 2013 01:24 AM
How to Apply the Shapley Value Formula? Aqil Economics 0 December 14th, 2012 11:31 AM
what are the limit questions whic you can't apply L'hopital' silvercats Calculus 5 June 26th, 2011 05:24 AM
Why does this apply to all tables except to a multiple of 3? DragonsMath Calculus 2 November 19th, 2010 07:34 AM
How To Figure Gross With a Certain Net Figure Al_Ch Elementary Math 2 May 21st, 2009 05:53 PM

Copyright © 2019 My Math Forum. All rights reserved.