Question

Find the limit

Answer 1

@biogirl Can you factor the numerator and denominator?

Answer 2

And that will be your solution

Answer 3

Whatdoes rhe top factor to? Would it be x^2-9/ (x+2)(x-1)

Answer 4

Use this for factoring numerator
\[a^2-b^2=(a+b)(a-b)\]
Can you try?

Answer 5

\[\lim_{x\rightarrow -3}{\frac{x^2-9}{x^2+2x-3}}=lim_{x\rightarrow -3}{\frac{(x+3)(x-3)}{(x-1)(x+3)}}=lim_{x\rightarrow -3}{\frac{x-3}{x-1}}=lim_{x=-3}{\frac{x-3}{x-1}}=\frac{6}{4}\]

Therefore, \(\lim_{x\rightarrow -3}{\frac{x^2-9}{x^2+2x-3}}=1.5\)