Answer:
B
Step-by-step explanation:
[tex]\lim_{x \to 5} \frac{x^2-7x+10}{x^2-14x+45}\\= \lim_{x \to 5} \frac{x^2-2x-5x+10}{x^2-5x-9x+45} \\= \lim_{x \to 5} \frac{x(x-2)-5(x-2)}{x(x-5)-9(x-5)} \\= \lim_{x \to 5}\frac{(x-2)(x-5)}{(x-5)(x-9)} \\= \lim_{x \to 5} \frac{x-2}{x-9} \\=\frac{5-2}{5-9} \\=-\frac{3}{4}\\also~f(5)=-3/4\\so~ it~ is~ continuous~ at ~x=5[/tex]
it does not exist at x=9.
so it is not continuous at x=9
