Question

Can someone solve this question ~

Answer 1

$\\ \sf{:}\Rrightarrow {\displaystyle{\lim_{x\to -3}}}\dfrac{x^2-9}{x^2+2x-3}$

$\\ \sf{:}\Rrightarrow {\displaystyle{\lim_{x\to -3}}}\dfrac{(x+3)(x-3)}{x^2+3x-x-3}$

$\\ \sf{:}\Rrightarrow {\displaystyle{\lim_{x\to -3}}}\dfrac{\cancel{(x+3)}(x-3)}{\cancel{(x+3)}(x-1)}$

$\\ \sf{:}\Rrightarrow {\displaystyle{\lim_{x\to -3}}}\dfrac{x-3}{x-1}$

• Apply limits

$\\ \sf{:}\Rrightarrow \dfrac{-3-3}{-3-1}$

$\\ \sf{:}\Rrightarrow \dfrac{-6}{-4}$

$\\ \sf{:}\Rrightarrow \dfrac{3}{2}$

Answer 2

Hope you could understand.

If you have any query, feel free to ask.