Given to solve,

`lim_(x->-2) (x^2-3x+10)/(x+2)`

as `x->-2` then the `(x^2-3x+10)/(x+2) =0/0` form

so upon applying the L 'Hopital rule we get the solution as follows,

as for the general equation it is as follows

`lim_(x->a) f(x)/g(x) is = 0/0` or `(+-oo)/(+-oo)` then by using the L'Hopital Rule we get the solution with the below form.

`lim_(x->a) (f'(x))/(g'(x))`

so , now evaluating

`lim_(x->-2) (x^2-3x+10)/(x+2)`

=`lim_(x->-2) ((x^2-3x+10)')/((x+2)')`

=`lim_(x->-2) (2x-3)/(1)`

now plugging the value of `x= -2` then we get

= `(2(-2)-3)`

= `-7`

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