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January 26th, 2019, 07:46 AM   #1
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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.
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January 26th, 2019, 08:42 AM   #2
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$\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x) \cdot \sin(\sqrt{x})}{\sqrt{x^3}}$

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.
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January 26th, 2019, 10:33 AM   #3
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You can also use L-Hoptial rule.
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January 26th, 2019, 11:04 AM   #4
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Quote:
Originally Posted by idontknow View Post
You can also use L-Hoptial rule.
You can't.

Successive differentiations just to increasing powers of $x$ in a denominator.
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January 26th, 2019, 11:10 AM   #5
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Thank you very much.
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January 26th, 2019, 11:21 AM   #6
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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
\[ \lim_{x \to 0^+} \frac{2x^{3/2} + \mathcal{O}(x^{5/2})}{x^{3/2}} = 2 \]
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January 26th, 2019, 11:45 AM   #7
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Quote:
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Successive differentiations just to increasing powers of $x$ in a denominator.
I mean to use differentation to the end $\displaystyle \frac{ln(1+2x)}{x}$ .
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January 27th, 2019, 02:36 AM   #8
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$\displaystyle \lim_{x\to0}(1+2x)^{1/x}=e^2$
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January 28th, 2019, 04:55 AM   #9
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Quote:
Originally Posted by greg1313 View Post
$\displaystyle \lim_{x\to0}(1+2x)^{1/x}=e^2$

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.
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January 29th, 2019, 10:15 AM   #10
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Quote:
Originally Posted by skeeter View Post
$\displaystyle \lim_{x \to 0^+} \dfrac{\ln(1+2x) \cdot \sin(\sqrt{x})}{\sqrt{x^3}}$

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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