The numerator can be factored as 2 x 2 + 11 x − 21 = ( 2 x − 3 ) ( x + 7 ) {\displaystyle 2x^{2}+11x-21=(2x-3)(x+7)} , and the dominator can be factored as x 2 + 8 x + 7 = ( x + 1 ) ( x + 7 ) {\displaystyle x^{2}+8x+7=(x+1)(x+7)} . Therefore we have,
lim x → − 7 2 x 2 + 11 x − 21 x 2 + 8 x + 7 = lim x → − 7 ( 2 x − 3 ) ( x + 7 ) ( x + 1 ) ( x + 7 ) = lim x → − 7 2 x − 3 x + 1 = ( 2 ⋅ − 7 − 3 ) ( − 7 + 1 ) = − 17 − 6 = 17 6 {\displaystyle {\begin{aligned}\lim _{x\to -7}{\frac {2x^{2}+11x-21}{x^{2}+8x+7}}=&\lim _{x\to -7}{\frac {(2x-3)(x+7)}{(x+1)(x+7)}}=\lim _{x\to -7}{\frac {2x-3}{x+1}}\\=&{\frac {(2\cdot -7-3)}{(-7+1)}}={\frac {-17}{-6}}={\frac {17}{6}}\end{aligned}}}
Answer: 17 6 {\displaystyle \color {blue}{\frac {17}{6}}}