Prove that x - √(x² - 1) ≈ 1/2x for large x?

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.

For large x, x² is much larger than 1

therefore x²–1 is the same as x²

therefore we have

x – √(x² – 1) ≈ x – √(x²) ≈ x – x ≈ 0

so your premise is incorrect.

try x = 1000

x – √(x² – 1) = 1000 – √(1e6 – 1) = 1000 – √999999 = 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 ≈ 1/(2x) ??

come back when you can be more specific.

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.