# `lim_(t->0) sin(3x)/(5x^3 - 4x)` Find the limit.

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### 1 Answer

`lim_(x->0) sin(3x)/(5x^3 - 4x)`

using L-Hopital's rule

we get

`lim_(x->0) sin(3x)/(5x^3 - 4x)`

`=lim_(x->0) (3*cos(3x))/(15x^2 - 4)`

Now as `x->0` , we get

`=(3*cos(3*0))/(15(0)^2 - 4)= -3/4`