Answer:
The answer is "Option a".
Step-by-step explannation:
[tex]\to \bold{ \lim_{n \to \5} \frac{(x^2+x-30)}{(x-5)} = \lim_{n \to \5} (x+6) }\\\\[/tex]
[tex]\to \lim_{n \to \5} \frac{(x^2+(6-5)x-30)}{(x-5)} = \lim_{n \to \5} (x+6) \\\\\to \lim_{n \to \5} \frac{(x^2+6x-5x-30)}{(x-5)} = \lim_{n \to \5} (x+6) \\\\\to \lim_{n \to \5} \frac{x(x+6)-5(x+6)}{(x-5)} = \lim_{n \to \5} (x+6) \\\\\to \lim_{n \to \5} \frac{(x+6) (x-5)}{(x-5)} = \lim_{n \to \5} (x+6) \\\\\to \lim_{n \to \5} (x+6) = \lim_{n \to \5} (x+6) \\\\\to 5+6 =5+6\\\\\to 11 =11[/tex]
that's why the choice a is correct.
