# `lim_(x->1/3)sin(3x-1)/(3x^2+2x-1)`

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The limit `lim_(x->1/3)sin(3x-1)/(3x^2+2x-1)` has to be determined.

Substituting x = `1/3` , gives the form `0/0` which is indeterminate. This allows the use of l'Hopital's Rule and the numerator and denominator can be replaced with their derivatives.

`lim_(x->1/3)sin(3x-1)/(3x^2+2x-1)`

= `lim_(x->1/3) ((sin(3x-1))')/((3x^2+2x-1)')`

= `lim_(x->1/3) (3*cos(3x-1))/(6x + 2)`

Substituting x = `1/3` gives `(3*1)/4 = 3/4`

**The required limit is 3/4**