Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $$a_n\xrightarrow[n\to\infty]{}+\infty, \ b_n\xrightarrow[n\to\infty]{}b>0 \ \Longrightarrow \ a_nb_n\xrightarrow[n\to\infty]{}+\infty.$$

share|cite|improve this answer
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 $$\lim_{v\to +\infty}\sqrt{v}\ln(1+\frac2{\sqrt{v}})=2$$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.