Calc Midterm: Solving the Limit Question (1-cosx)/(x^2) | 0.5 or Nonexistent?

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SUMMARY

The limit as X approaches 0 of (1-cosx)/(x^2) evaluates to 0.5. This conclusion is reached by multiplying the numerator and denominator by the conjugate (1+cosx) and factoring out (sinx/x) twice. Some participants incorrectly argue that the limit does not exist due to the denominator approaching zero, but the correct application of limits shows that the limit is indeed 0.5.

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ok, so today we had our calc midterm, and on it was this question:

the limit as X->0 of (1-cosx)/(x^2)

what i did was multiply the top and bottom with the conjugate of (1-cosx) which is (1+cosx). then i managed to factor out (sinx/x) twice. since (sinx/x) is just one, iarrived at the answer of 0.5 for the limit of this question.

some of my friends tell me that limit does not exist, because the (X^2) would evaluate to zero and you can't have a zero in the denominator.

so who's right?
 
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you are right. [tex]\lim_{x\rightarrow 0 } \frac{1-\cos^{2}x}{x^{2}(1+\cos x)} = \frac{\sin^{2}x}{x^{2}(1+\cos x)} = \frac{\sin x}{x} \frac{\sin x}{x}\frac{1}{1+\cos x} = \frac{1}{2}[/tex]
 
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yes! thank you.
 

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