#lim x->o (tan( x/3)/x)# ???

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#lim x->o (tan( x/3)/x)#

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Answer 1 · 2018-06-20T13:17:34

We have to use that #lim_(x->0)sin x/x =1# to solve the indetermination

let's manipulate the expression a little bit first:

#lim_(x->0) (tan (x/3)/x) =lim_(x->0) ((sin (x/3)/cos(x/3))/x) = lim_(x->0) (sin (x/3)/(cos(x/3)x)) # and now dividing by 3

#lim_(x->0) (tan (x/3)/x) = lim_(x->0) ((sin (x/3)/3)/(cos(x/3)(x/3))) = lim_(x->0) ((sin (x/3)/(x/3))* 1/(cos(x/3)*3)) #

But we know that:

#lim_(x->0) (sin (x/3)/(x/3)) =1#, and that

#lim_(x->0) 1/cos(x/3) =1 #, so the limit is

#lim_(x->0) (tan (x/3)/x) = lim_(x->0) ((sin (x/3)/3)/(cos(x/3)(x/3))) = lim_(x->0) ((sin (x/3)/(x/3))* 1/(cos(x/3)*3)) = 1*1*1/3= 1/3#