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How do you find the limit of #(costheta-1)/sintheta# as #theta->0#?

Answer 1

Aurora Alvarado

#Lt_(theta->0)(costheta-1)/sintheta=0#

#Lt_(theta->0)(costheta-1)/sintheta#

= #Lt_(theta->0)(costheta-1)/sinthetaxx(costheta+1)/(costheta+1)#

= #Lt_(theta->0)(cos^2theta-1)/(sinthetaxx(costheta+1))#

= #Lt_(theta->0)(-sin^2theta)/(sinthetaxx(costheta+1))#

= #Lt_(theta->0)(-sintheta)/(costheta+1)#

= #(-sin0)/(cos0+1)#

= #-0/2=0# graph{(cosx-1)/sinx [-10, 10, -5, 5]}

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Answer 2

Adam Adkins

To find the limit of (cos(theta) - 1)/sin(theta) as theta approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (-sin(theta))/cos(theta). Evaluating this expression as theta approaches 0, we have (-sin(0))/cos(0) = 0/1 = 0. Therefore, the limit of (cos(theta) - 1)/sin(theta) as theta approaches 0 is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #(costheta-1)/sintheta# as #theta-&gt;0#?

How do you find the limit of #(costheta-1)/sintheta# as #theta->0#?