Factor the numerator and denominator. If you can cancel out an $(x−2)$, then the limit exists. If not, then it DNE.

You could factor the denominator. Or you could multiply the top and bottom by the conjugate of $(\sqrt{x}-5)$

The $(x−2)$ in the denominator will not cancel out. This means that there is a vertical asymptote at $x = 2$. How can we determine if the asymptote approaches $+∞$ or $−∞$ as $x → 2^−$ from the left?

Choose a value that is to left of $x = 2$, and evaluate. Note: we only need to determine if each term is positive or negative. Choose $x → 1.9$:
$\lim\limits_{x\rightarrow 1.9}\frac{(2x+1)(x-5)^{6}}{(x-2)^{7}\sqrt{9-x}}=\lim\limits_{x\rightarrow 1.9}\frac{(+)(-)^{6}}{(-)^{7}(\sqrt{+})}$
$=\lim\limits_{x\rightarrow 1.9}\frac{(+)(+)}{(-)(+)}= -\text{ value}$

DNE but $y → −∞$.

This is a limit $x→∞$. That means we are looking at end behavior. Is there a horizontal asymptote, or do the y-values approach $∞$ or $−∞$?

Consider only the terms with the highest power in the numerator and the denominator. Also consider their coefficients, and their signs.