Evaluate a limit
- Thread starter LoudBar
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D
Deleted member 4993
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D
Deleted member 4993
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...... of doing WHAT?
skeeter
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I assume you're trying to determine [MATH]\lim_{n \to \infty} \{ x_n \}[/MATH]
note ...
[MATH](a-b)(a^2+ab+b^2) = a^3-b^3[/MATH]
let [MATH]a = \sqrt[3]{1+\dfrac{2}{n}}[/MATH] and [MATH]b = 1[/MATH]
... multiply your expression by [MATH]\dfrac{a^2+ab+b^2}{a^2+ab+b^2}[/MATH]
note ...
[MATH](a-b)(a^2+ab+b^2) = a^3-b^3[/MATH]
let [MATH]a = \sqrt[3]{1+\dfrac{2}{n}}[/MATH] and [MATH]b = 1[/MATH]
... multiply your expression by [MATH]\dfrac{a^2+ab+b^2}{a^2+ab+b^2}[/MATH]
Yeah, i tried it and idk where is my mistake or what i suppose to do next
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D
Deleted member 4993
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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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Steven G
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Sad, that is all I have to say.
pka
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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.