Question #ca2da

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Answer 1 · 2016-06-21T20:52:47

#9/2#

this is one i think you can hit very hard with L'Hopital's rule

again

Wikipedia

so
#lim_(x->0)(3x-sin3x)/(x^3) = lim_(x->0)(3-3cos3x)/(3x^2) = lim_(x->0)(9sin3x)/(6x)#

but this is where we have to stop because if we re-write it as

# lim_(x->0)(9sin3x)/(6x) = lim_(x->0) 9/2 (sin3x)/(3x)#

and make the sub #p = 3x# , then we have:

# lim_(p->0) 9/2 (sin p)/(p) = 9/2 lim_(p->0) (sin p)/(p)#

we cannot use L'Hopital on that limit, because that limit is used in the derivation of #(sin x)'#

but we know from Squeeze Theorem, for example, that # lim_(p->0) (sin p)/(p) = 1#

so the limit is #9/2#