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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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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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  • $\begingroup$ Thanks. So this limit is $$2\cdot\sqrt{n}\cdot 2=\infty$$? $\endgroup$ – Favolas Dec 28 '12 at 18:41
  • $\begingroup$ @Favolas: Yes the limit is $\infty$. To be formal you should say: since $\lim2\sqrt{n}=\infty$ and $\lim \cdots =2 \Rightarrow \lim 2\sqrt{n} \cdots =\infty$. $\endgroup$ – P.. Dec 29 '12 at 10:58
  • $\begingroup$ Thanks for your explanation $\endgroup$ – Favolas Dec 29 '12 at 12:00
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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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  • $\begingroup$ Thanks but I believe you forgot the square root $\endgroup$ – Favolas Dec 28 '12 at 18:34
  • $\begingroup$ @Favolas I didn't forget any square root. That's why I said : "Substitute some more and you'll get it" $\endgroup$ – Nameless Dec 28 '12 at 18:35
  • $\begingroup$ Ok. Sorry. I had problems with the square root $\endgroup$ – Favolas Dec 28 '12 at 18:43

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