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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.

2 Answers 2

5

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 for your e$x$planation2012-12-29
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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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    Ok. Sorry. I had problems with the square root2012-12-28