Question

plz guys help me..
limits and continuity​

Answer 1

Answer:

       13 / 8

Step-by-step explanation:

$\displaystyle\lim_{x\rightarrow4}\left(\frac{2x^2-4x-24}{x^2-16}-\frac1{4-x}\right)\\\\=\lim_{x\rightarrow4}\left(\frac{2x^2-4x-24}{x^2-16}+\frac{x+4}{x^2-16}\right)\\\\=\lim_{x\rightarrow4}\left(\frac{2x^2-3x-20}{x^2-16}\right)\\\\=\lim_{x\rightarrow4}\frac{(x-4)(2x+5)}{(x-4)(x+4)}\\\\=\lim_{x\rightarrow4}\frac{2x+5}{x+4}\\\\=\frac{8+5}{4+4}=\frac{13}8$