`lim_(x->1) ((x^4-3x^2+2)/(x^3-5x^2+3x+1))`

Explanation

Given,

`lim_(x->1)(x^4−3x^2+2)/(x^3−5x^2+3x+1)`

`=lim_(x->1)(1^4−3xx1^2+2)/(1^3−5xx1^2+3xx1+1)`

`=0/0`


Thus, applying l-Hospital rule,


`lim_(x->1)(x^4−3x^2+2)/(x^3−5x^2+3x+1)`

...