How do I solve this... L'Hopitals seems to go in an loop?

Question

lim x-> 0 of #sin^3x/sinx^3# =?

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Answer 1 · 2018-02-23T21:30:50

#1#

Assuming the basic result

#lim_(x->0) sinx/x = 1# (Textbook result) we have

#(sin^3x)/(sin x^3) =( x^3/x^3)(sin^3x)/(sin x^3) = (sinx/x)^3(x^3/sin x^3)#

then

#lim_(x->0)(sin^3x)/(sin x^3)=lim_(x->0)(sinx/x)^3 lim_(x->0)(x^3/sin x^3) = 1 xx 1 = 1#

Answer 2 · 2018-02-23T21:33:09

I'm not pretty sure but i hope this helps...

#lim_(x rarr0) (sin^3x)//sin x^3#
Multiplying denominator with #x^3#,we have,
#lim_(xrarr0)[{sin^3x}//x^3 ]xx1//{(sinx^3)//x^3}#
thus #(1)^3xx1#=1