Limit withoul using the L'Hopital's rule

[Shaa-Maan]

Hey folks,

New member here (not a native english speaker, but i'll try my best)

I've started attending a university just this year, and now i'm studying for an exam, and i've encountered this ''tricky'' limit in one of my books and it says specifically not to use the L'Hopital's rule.(sorry, but i'll write this in plain text)

lim n to infinity [n/2 (((1+2/n)^(1/3))-1)]

so i've tried pretty much everything, substituting the 3rd root, playing around with it trying to get something to ''cancel out'' but it seems like everything i do i always end up moving in a circle, always coming back to where i started.

according to my book, symbolab, wolfram, and numerous other sources, the limit = 1/3.

anyway, i'm sure there is just something wrong with how i look at it, and i hope some of you good people could help me out here.

btw, sorry if it's hard to understand but as i said i'm not a native speaker.

thx for help in advance

 

[stapel]

Just to be sure, your limit is as follows, right?

. . . . .\(\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,\)\(\displaystyle \left(\dfrac{n}{2}\right)\, \left(\sqrt[3]{\strut 1\, +\,\dfrac{2}{n}\,}\, -\, 1\right)\)

When you reply, please include at least one of your efforts. Thank you!

 

[Shaa-Maan]

Just to be sure, your limit is as follows, right?

. . . . .\(\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,\)\(\displaystyle \left(\dfrac{n}{2}\right)\, \left(\sqrt[3]{\strut 1\, +\,\dfrac{2}{n}\,}\, -\, 1\right)\)

When you reply, please include at least one of your efforts. Thank you!

yup that's the limit. Since I don't know how to write this here so it won't look like trash, can i attach a photo of my (hand written) work?

 

[stapel]

can i attach a photo of my (hand written) work?

Sure. That'll work. Thank you!

 

[Deleted member 4993]

\(\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,\)\(\displaystyle \left(\dfrac{n}{2}\right)\, \left(\sqrt[3]{\strut 1\, +\,\dfrac{2}{n}\,}\, -\, 1\right)\)

Have you covered Taylor's series expansion?

Do you know how to expand

(1 + x)^1/3

using Taylor's series?

 

[Shaa-Maan]

Sure. That'll work. Thank you!

oh never mind, i figured it out.

Like i said i looked at it completely wrong, and what i had to do was so obvious, that i didn't try it at first, because i thought it's too good to be true

(1+2/n)^1/3 = t and multiply by (t^2 + t + 1)/(t^2 + t + 1)

anyway i'm attaching the limit for other people to see how blind i was

nevertheless, thx for being willing to help. I appreciate it.

 

[Shaa-Maan]

\(\displaystyle \displaystyle \lim_{n\, \rightarrow\, \infty}\,\)\(\displaystyle \left(\dfrac{n}{2}\right)\, \left(\sqrt[3]{\strut 1\, +\,\dfrac{2}{n}\,}\, -\, 1\right)\)

Have you covered Taylor's series expansion?

Do you know how to expand

(1 + x)^1/3

using Taylor's series?

of course, but this exercise was about doing it "the old fashioned way". If given the chance, no doubt I'd use L'Hospital or Taylor to get it out of the way