`lim_(x->1) (x^11-1)/(x^4-1)`

To solve, plug-in x = 1.

`lim_(x->1) (x^11-1)/(x^4-1) = (1^11-1)/(1^4-1) =0/0`

Since the result is indeterminate, to determine the limit of the function as x approaches 1, apply L'Hopital's Rule. To do so, take the derivative of the numerator and denominator.

`lim_(x->1) (x^11-1)/(x^4-1) = lim_(x->1) ((x^11-1)')/((x^4-1)') = lim_(x->1)...

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`lim_(x->1) (x^11-1)/(x^4-1)`

To solve, plug-in x = 1.

`lim_(x->1) (x^11-1)/(x^4-1) = (1^11-1)/(1^4-1) =0/0`

Since the result is indeterminate, to determine the limit of the function as x approaches 1, apply L'Hopital's Rule. To do so, take the derivative of the numerator and denominator.

`lim_(x->1) (x^11-1)/(x^4-1) = lim_(x->1) ((x^11-1)')/((x^4-1)') = lim_(x->1) (11x^10)/(4x^3) = lim_(x->1) (11x^7)/4`

And, plug-in x=1.

`= (11*1^7)/4 = 11/4`

**Therefore, `lim_(x->1) (x^11-1)/(x^4-1)=11/4` .**