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 January 26th, 2019, 06:46 AM #1 Newbie   Joined: Jan 2019 From: Athens Posts: 5 Thanks: 0 Help for calculating a limit I am trying to calculate the following limit $\displaystyle \lim_{x\to0+}\frac{\ln(1+2x)\sin(\sqrt{x})}{\sqrt{ x^3}}$ but no luck. Can anybody help me with this? Last edited by greg1313; January 27th, 2019 at 01:44 AM. January 26th, 2019, 07:42 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,005 Thanks: 1588 $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x) \cdot \sin(\sqrt{x})}{\sqrt{x^3}}$ note $\sqrt{x^3} = \sqrt{x^2} \cdot \sqrt{x} = x \cdot \sqrt{x}$ for $x>0$ $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x)}{x} \cdot \lim_{x \to 0^+} \dfrac{\sin(\sqrt{x})}{\sqrt{x}}$ $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x)}{x} \cdot 1$ from here, I recommend using the series expansion about zero for $\ln(1+2x)$ , simplifying the resulting quotient, and then determine the limit. Thanks from topsquark and romsek January 26th, 2019, 09:33 AM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 606 Thanks: 88 You can also use L'Hôpital rule. Last edited by skipjack; March 26th, 2019 at 12:36 AM. January 26th, 2019, 10:04 AM   #4
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Quote:
 Originally Posted by idontknow You can also use L'Hôpital rule.
You can't.

Successive differentiations just to increasing powers of $x$ in a denominator.

Last edited by skipjack; March 26th, 2019 at 12:37 AM. January 26th, 2019, 10:10 AM #5 Newbie   Joined: Jan 2019 From: Athens Posts: 5 Thanks: 0 Thank you very much. January 26th, 2019, 10:21 AM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Just expand to first order, compose and multiply. Start with $\ln(1+2x) = 2x + \mathcal{O}(x^2)$ and $\sin(\sqrt{x}) = \sqrt{x} + \mathcal{O}(x^{3/2})$ so when you multiply you have $\ln(1+2x)\sin(\sqrt{x}) = 2x^{3/2} + \mathcal{O}(x^{5/2})$. Obviously if you divide by $x^{3/2}$ and take $x \to 0$ you get $\lim_{x \to 0^+} \frac{2x^{3/2} + \mathcal{O}(x^{5/2})}{x^{3/2}} = 2$ Thanks from topsquark and idontknow January 26th, 2019, 10:45 AM   #7
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Quote:
 Originally Posted by romsek Successive differentiations just to increasing powers of $x$ in a denominator.
I mean to use differentiation to the end $\displaystyle \frac{\ln(1+2x)}{x}$ .

Last edited by skipjack; March 26th, 2019 at 12:38 AM. January 27th, 2019, 01:36 AM #8 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond $\displaystyle \lim_{x\to0}(1+2x)^{1/x}=e^2$ Thanks from topsquark and v8archie January 28th, 2019, 03:55 AM   #9
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Quote:
 Originally Posted by greg1313 $\displaystyle \lim_{x\to0}(1+2x)^{1/x}=e^2$

If you're wondering how it's done (using $\lim_{x\to0}(1+x)^{1/x}=e$)...

$\displaystyle u=ax$

$\displaystyle \lim_{x\to0}(1+ax)^{1/x}=\lim_{u\to0}(1+u)^{a/u}=\left[\lim_{u\to0}(1+u)^{1/u}\right]^a=e^a$

where $a$ is a non-zero constant. January 29th, 2019, 09:15 AM   #10
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Quote:
 Originally Posted by skeeter $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x) \cdot \sin(\sqrt{x})}{\sqrt{x^3}}$ note $\sqrt{x^3} = \sqrt{x^2} \cdot \sqrt{x} = x \cdot \sqrt{x}$ for $x>0$ $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x)}{x} \cdot \lim_{x \to 0^+} \dfrac{\sin(\sqrt{x})}{\sqrt{x}}$ $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x)}{x} \cdot 1$ from here, I recommend using the series expansion about zero for $\ln(1+2x)$ , simplifying the resulting quotient, and then determine the limit.
Good grief. Why all the contortions?

By L'Hôpital's rule:

lim log(1+2x)/x = lim 2/(1+2x) = 2

Last edited by skipjack; March 26th, 2019 at 12:40 AM. Tags calculating, limit 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 jnicholes Math 5 January 25th, 2019 03:23 PM zylo Calculus 13 May 31st, 2017 12:53 PM samdon123 Calculus 3 October 18th, 2016 08:00 PM expertalmost Advanced Statistics 0 February 28th, 2014 05:03 AM ricsi046 Real Analysis 1 October 7th, 2013 09:02 AM

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