# Evaluate a limit

#### LoudBar

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

#### firemath

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

Staff member

#### firemath

##### Full Member
S/he wants the process.....?

Staff member

#### firemath

##### Full Member
Moving from non-bracketed problem 15 to the $$\displaystyle 1/3$$ in brackets? Maybe?

#### skeeter

##### Elite Member
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

##### New member
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

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#### Subhotosh Khan

##### Super Moderator
Staff member
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

##### New member
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!

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#### Jomo

##### Elite Member
Moving from non-bracketed problem 15 to the $$\displaystyle 1/3$$ in brackets? Maybe?
Sad, that is all I have to say.

#### lookagain

##### Elite Member
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.