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Prove that x - √(x² - 1) ≈ 1/2x for large x?
I'd like to see how this can easily be determined.
Eugene, d'oh! Why didn't I think of that?
Rita the Dog, yes, that's the series for large x, by finding the Maclauren for (1/x) - √((1/x)² - 1) for x = 0.
Anyway, I wasted too much time in coming around to this in answering your question about that x = Tan(x) series. I had just assumed falsely that this expression would be more like 1/x² instead of the correct 1/x. So I learned something.
3 Answers
- EugeneLv 78 years agoFavorite Answer
Let f(x) = x - √(x^2 - 1) and g(x) = 1/(2x). We want to show lim(x -> ∞) f(x)/g(x) = 1.
lim(x -> ∞) f(x)/g(x)
= lim [x - √(x^2 - 1)]/(1/2x)
= lim 2x/(x + √(x^2 - 1), by rationalizing the numerator
= lim 2/(1 + √(1 - 1/x^2)) = 2/(1 + 1)
= 1.
- billrussell42Lv 78 years ago
For large x, x² is much larger than 1
therefore x²–1 is the same as x²
therefore we have
x – â(x² – 1) â x – â(x²) â x – x â 0
so your premise is incorrect.
try x = 1000
x – â(x² – 1) = 1000 – â(1e6 – 1) = 1000 – â999999 = 1000 – 999.9995 = 0.0005
or approximately 0 compared to 1000
(1/2)x = 500 which is not close to 0.0005
BUT wait, perhaps you mean â 1/(2x) ??
come back when you can be more specific.
- Rita the dogLv 78 years ago
The series expansion is 1/2x + 1/8x^3 + 1/16x^5 + 5/128x^7 + ... but I´ll leave a real proof to someone else.
