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How do you find the limit of #(tan^3 (2x))/ x^3# as x approaches 0?

Answer 1

Paisley Johnson

#8#

#lim_(x->0) tan^3(2x)/x^3=8lim_(x->0) tan^3(2x)/(8x^3)=8lim_(x->0) tan^3(2x)/(2x)^3=#

#=8lim_(x->0) (tan(2x)/(2x))^3=8(lim_(x->0) (tan2x)/(2x))^3=#

#=8(lim_(x->0) (sin2x)/(2x) 1/cos(2x))^3 =# #=8(lim_(x->0) (sin2x)/(2x) lim_(x->0) 1/cos(2x))^3 = A#

#2x=t => x->0 =>t->0#

#A=8(lim_(t->0) (sint)/(t) lim_(t->0) 1/cost)^3 = 8 * (1*1)^3 = 8#

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Answer 2

Christopher Agresta

To find the limit of (tan^3 (2x))/ x^3 as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get 6tan^2(2x)sec^2(2x) and 3x^2. Evaluating these expressions at x=0, we find that the limit 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.