Evaluate a limit

[LoudBar]

Please help. The answer is in the brackets, but how to get there...

 

[firemath]

What have you tried? Where are you stuck? If you tell us these things, we can help you better.

 

[Subhotosh Khan]

Please help. The answer is in the brackets, but how to get there...

View attachment 21654

Your image does not have a question!

What is it that you are trying to "FIND"?

 

[firemath]

S/he wants the process.....?

 

[Subhotosh Khan]

S/he wants the process.....?

...... of doing WHAT?

 

[firemath]

Moving from non-bracketed problem 15 to the \(\displaystyle 1/3\) in brackets? Maybe?

 

[skeeter]

I assume you're trying to determine \(\displaystyle \lim_{n \to \infty} \{ x_n \}\)

note ...

\(\displaystyle (a-b)(a^2+ab+b^2) = a^3-b^3\)

let \(\displaystyle a = \sqrt[3]{1+\dfrac{2}{n}}\) and \(\displaystyle b = 1\)

... multiply your expression by \(\displaystyle \dfrac{a^2+ab+b^2}{a^2+ab+b^2}\)

 

[LoudBar]

I assume you're trying to determine \(\displaystyle \lim_{n \to \infty} \{ x_n \}\)

note ...

\(\displaystyle (a-b)(a^2+ab+b^2) = a^3-b^3\)

let \(\displaystyle a = \sqrt[3]{1+\dfrac{2}{n}}\) and \(\displaystyle b = 1\)

... multiply your expression by \(\displaystyle \dfrac{a^2+ab+b^2}{a^2+ab+b^2}\)

Yeah, i tried it and idk where is my mistake or what i suppose to do next

 

[Subhotosh Khan]

Yeah, i tried it and idk where is my mistake or what i suppose to do next

Your work is correct.

Now take the limit - the denominator should go to 3 as n \(\displaystyle \to \infty \) ..... [& 1/n \(\displaystyle \to \) 0]

 

[LoudBar]

Your work is correct.

Now take the limit - the denominator should go to 3 as n \(\displaystyle \to \infty \) ..... [& 1/n \(\displaystyle \to \) 0]

I think i got it, thank You a lot for helping me!

 

[Jomo]

Moving from non-bracketed problem 15 to the \(\displaystyle 1/3\) in brackets? Maybe?

Sad, that is all I have to say.

 

[pka]

Please help. The answer is in the brackets, but how to get there...

View attachment 21654

LOOK HERE

 

[lookagain]

Please help. The answer is in the brackets, but how to get there...

View attachment 21654

Here are a couple of approximations that may help you sometimes:

\(\displaystyle \sqrt{a + b} \ \approx \ \sqrt{a} + \dfrac{1}{2}b\)

\(\displaystyle \sqrt[3]{a + b} \ \approx \ \sqrt[3]{a} + \dfrac{1}{3}b\)

For this problem, a = 1 and b = 2/n for the cube root approximation.

The expression then is approximated by

\(\displaystyle \dfrac{n}{2}\bigg(1 + \dfrac{2}{3n} \ - \ 1 \bigg)\)

Simplify that and then take the limit as n approaches infinity.