Using the L’Hopital’s Rule, evaluate lim→0 1 − cos 2 [4

Question

using L'Hopitals rule, evaluate  lim 1-cos 2

                                                   x→0

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Answer 1

My guess is that you mean:

lim_x→0 (1-cos(2x))/4x or something similar.

If this is the case 1-cos(2x) can be written approximately:

(1-(1-(2x)²/2))/4x=2x²/4x=x/2,

which becomes zero as x→0, so the limit is zero. The important thing is that the power of x in the numerator is higher than that in the denominator so the numerator predominates when calculating the limit. If the denominator had been x⁴, rather than 4x, for example, then the power of x in the denominator would take precedence over that in the numerator and the limit would be +∞ (positive infinity). The constant coefficients in this case are irrelevant. What matters is where the higher powers of x sit.

Using l'Hôpitals's Rule we would differentiate the numerator and denominator.

For lim_x→0 (1-cos(2x))/4x we'd get 2sin(2x) and 4, so we can write lim_x→0 2sin(2x)/4→0, making the limit zero, because the numerator is zero when x=0 and the denominator is a positive constant.

For lim_x→0 (1-cos(2x))/x⁴ we'd get 2sin(2x)/4x³ which doesn't help to find the limit since both denominator and numerator go to zero in the limit, and l'Hôpitals's Rule doesn't really help. You will need to apply the rule to the original question.