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