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Imagine I have this limit:

$$\lim_{x\to 0}\frac{\ln(1+2x)}x$$

Using the L'Hospital's rule the result is $2$.

Using this result is it possible to calculate

$$\lim_{n\to \infty}\ n\ln\bigg(1+\frac{4}{\sqrt{n}}\bigg) \quad ?$$

Sorry if this is an easy question, but many years have passed since I've learned calculus.

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2 Answers 2

up vote 5 down vote accepted

Note that $$n\ln\bigg(1+\frac{4}{\sqrt{n}}\bigg)=2\cdot\sqrt{n}\cdot\dfrac{\ln\bigg(1+2\cdot\frac{2}{\sqrt{n}}\bigg)}{\frac{2}{\sqrt{n}}}.$$ Now use that $$\lim_{x\to 0}\frac{\ln(1+2x)}x=2$$ and the fact that $$a_n\xrightarrow[n\to\infty]{}+\infty, \ b_n\xrightarrow[n\to\infty]{}b>0 \ \Longrightarrow \ a_nb_n\xrightarrow[n\to\infty]{}+\infty.$$

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Thanks. So this limit is $$2\cdot\sqrt{n}\cdot 2=\infty$$? –  Favolas Dec 28 '12 at 18:41
    
@Favolas: Yes the limit is $\infty$. To be formal you should say: since $\lim2\sqrt{n}=\infty$ and $\lim \cdots =2 \Rightarrow \lim 2\sqrt{n} \cdots =\infty$. –  P.. Dec 29 '12 at 10:58
    
Thanks for your explanation –  Favolas Dec 29 '12 at 12:00

Hint: If you substitute $u=\frac1x$ then $$\lim_{u\to +\infty}u\ln(1+\frac2u)=2$$ This looks kind of like your limit. Substitute some more and you'll get it. For example substitute $v=u^2$. Then $$\lim_{v\to +\infty}\sqrt{v}\ln(1+\frac2{\sqrt{v}})=2$$

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Thanks but I believe you forgot the square root –  Favolas Dec 28 '12 at 18:34
    
@Favolas I didn't forget any square root. That's why I said : "Substitute some more and you'll get it" –  Nameless Dec 28 '12 at 18:35
    
Ok. Sorry. I had problems with the square root –  Favolas Dec 28 '12 at 18:43

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