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Limit as x goes zero of (1 - cosx)/x^2

Hi Roy,

I'm sure you've found that direct substitution results in the 0/0 form which is indeterminate. If you're familiar with calculus we could use the L'Hopital Rule (you can read about it) which says that we can differentiate the numerator and denominator of the function and do the substitution again to find the limit.

So,

lim x→0 [(1 - cosx)/x²] = lim x→0 (sinx/2x)

Substitution still results in 0/0 and so we apply the rule again.

lim x→0 (sinx/2x) = lim x→0 (cosx/2)

and substitution gives us a value of 1/2 for the limit of (1 - cosx)/x² as x approaches 0.

This approach might become tedious at times and in general it's better to use series expansion, if you're familiar with that. Around x = 0, we can expand thus;

cosx = 1 - x²/2! + x^4/4! - x^6/6! + ... (infinite terms in that fashion which we might not always need or show)

and so

1 - cosx = x²/2! - x^4/4! + x^6/6! - ...

(1 - cosx)/x² = 1/2! - x²/4! + x^4/6! - ...

Direct substitution of the limit here quickly gives us the value of 1/2, as before.

Regards

Calculus

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