Lim x->0 (tan x- sin x)/x^3=?

Question

a. 0
b. 1
c. 1/2
d. -1
e. -1/2

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Answer 1 · 2017-11-09T00:14:27

#lim_(x->0) (tanx-sinx)/x^3 = 1/2#

Transform the function in this way:

#(tanx-sinx)/x^3 = 1/x^3(sinx/cosx-sinx)#

#(tanx-sinx)/x^3 = 1/x^3((sinx-sinxcosx)/cosx)#

#(tanx-sinx)/x^3 = sinx/x^3(1-cosx)/cosx#

#(tanx-sinx)/x^3 = (sinx/x)((1-cosx)/x^2)( 1/cosx)#

We can use now the well known trigonometric limit:

#lim_(x->0) sinx / x = 1#

and using the trigonometric identity:

#sin^2alpha = (1-cos2alpha)/2#

we have:

#lim_(x->0) (1-cosx)/x^2 = lim_(x->0) (2sin^2(x/2))/x^2 = 1/2 lim_(x->0) (sin(x/2)/(x/2))^2 = 1/2#

While the third function is continuous so:

#lim_(x->0) 1/cosx = 1/1 = 1#

and we can conclude that:

#lim_(x->0) (tanx-sinx)/x^3 = lim_(x->0) (sinx/x)((1-cosx)/x^2)( 1/cosx) = 1 xx 1/2 xx 1 = 1/2#

graph{(tanx-sinx)/x^3 [-1.25, 1.25, -0.025, 1]}