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 January 26th, 2019, 07:46 AM #1 Newbie   Joined: Jan 2019 From: Athens Posts: 3 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 02:44 AM.
 January 26th, 2019, 08:42 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,818 Thanks: 1462 $\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x) \cdot \sin(\sqrt{x})}{\sqrt{x^3}}$ 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, 10:33 AM #3 Senior Member   Joined: Dec 2015 From: iPhone Posts: 387 Thanks: 61 You can also use L-Hoptial rule.
January 26th, 2019, 11:04 AM   #4
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Quote:
 Originally Posted by idontknow You can also use L-Hoptial rule.
You can't.

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

 January 26th, 2019, 11:10 AM #5 Newbie   Joined: Jan 2019 From: Athens Posts: 3 Thanks: 0 Thank you very much.
 January 26th, 2019, 11:21 AM #6 Senior Member   Joined: Sep 2016 From: USA Posts: 559 Thanks: 324 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, 11:45 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 differentation to the end $\displaystyle \frac{ln(1+2x)}{x}$ .

 January 27th, 2019, 02:36 AM #8 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,913 Thanks: 1112 Math Focus: Elementary mathematics and beyond $\displaystyle \lim_{x\to0}(1+2x)^{1/x}=e^2$ Thanks from topsquark and v8archie
January 28th, 2019, 04:55 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, 10:15 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'Hospital's rule:

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

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