\displaystyle\frac{{3}}{{4}} Explanation: \displaystyle\lim_{{{x}\to-{3}}}\frac{{{x}^{{2}}-{9}}}{{{x}^{{2}}-{2}{x}-{15}}} By factoring out the numerator and the denominator, \displaystyle=\lim_{{{x}\to-{3}}}\frac{{\cancel{{{\left({x}+{3}\right)}}}{\left({x}-{3}\right)}}}{{\cancel{{{\left({x}+{3}\right)}}}{\left({x}-{5}\right)}}}=\frac{{-{3}-{3}}}{{-{3}-{5}}}=\frac{{-{6}}}{{-{8}}}=\frac{{3}}{{4}} ...

\require{cancel} Note that both the numerator and denominator evaluate to 0 at x = -3. That means we have an indeterminate limit, which simply means more work needs to be done. Factor ...

HINT: \frac{x^2+x-6}{x^2-9}-\frac56=\frac{6x^2+6x-36-(5x^2-45)}{6(x^2-9)}=\frac{x^2+6x+9}{6(x^2-9)}=\frac{(x+3)^2}{6(x-3)(x+3)}

You can see if you solve for x=1 that you get a fraction in the form of \dfrac{0}{0} which is indeterminate. When you get into a situation like this you can apply L'Hôpital's rule - Wikipedia. ...

Let \epsilon>0 be given. We have to find a number \delta>0 such that \left|\frac{x^2-5x}{x^2+2}+1\right|<\epsilon whenever |x-2|<\delta. But, as Andre notes, \left|\frac{x^2-5x}{x^2+2}+1\right|=\left|\frac{2x^2-5x+2}{x^2+2}\right|=|x-2|\left|\frac{2x-1}{x^2+1}\right|. ...

Note that \lim_{x\to -3} \frac{(x^2-9)^2}{x^2+\color{red}6x+9} We can solve in this way \lim_{x\to -3} \frac{(x^2-9)^2}{(x+3)^2}=\lim_{x\to -3} \frac{(x-3)^2(x+3)^2}{(x+3)^2}=\lim_{x\to -3} (x-3)^2=36