\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

This is correct so far, but you should go on, somehow introducing the definition of f'. Briefly, it goes like t:=f^{-1}(x+h)-f^{-1}(x), we need that t\to 0 as h\to 0, and then consider y:=f^{-1}(x) ...

5 Undefined 0 2

To factor the numerator of the first limit, all you have to do is find the roots of the quadratic (use the quadratic formula, if everything else fails). Here, the two roots are 2 and 3, so x^2-5x+6 = (x-2)(x-3) ...