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How do you find the Limit of #(sin^3 x )/ (sin x - tan x)# as x approaches 0?

Answer 1

#-2#

#(sin^3 x )/ (sin x - tan x)=(sin x cdot sin^2x)/(sinx(1-1/cosx))=sin^2x/(1-1/cosx)# #=((1-cos^2x)cosx)/(cosx-1)=-((1-cosx)(1+cosx)cosx)/(1-cosx)#

#=-(1+cosx)cosx#

then

#lim_{x->0}(sin^3 x )/ (sin x - tan x)=lim_{x->0}(-(1+cosx)cosx) = -2#

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Answer 2

Ezekiel Alexander

To find the limit of (sin^3 x) / (sin x - tan x) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get:

d/dx (sin^3 x) = 3sin^2 x * cos x d/dx (sin x - tan x) = cos x - sec^2 x

Now, we can evaluate the limit by plugging in x = 0 into the derivatives:

lim (x->0) (3sin^2 x * cos x) / (cos x - sec^2 x)

Plugging in x = 0, we get:

lim (x->0) (3sin^2 0 * cos 0) / (cos 0 - sec^2 0) = 0 / (1 - 1) = 0

Therefore, the limit of (sin^3 x) / (sin x - tan x) as x 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.