Evaluate the limit, using L'Hôpital's Rule if necessary. $$ \lim _{x \rightarrow 4^{-}} \frac{\sqrt{16-x^{2}}}{x-4} $$

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Calculate the limit $$ \lim_{x \rightarrow 4^{-}} \frac{\sqrt{16-x^{2}}}{x-4} $$.
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
  $$\lim _{x\rightarrow 4^{-}}\left(\frac{\sqrt{16-x^{2}}}{x-4}\right)$$
- step1: Use the L'Hopital's rule:
  $$\lim _{x\rightarrow 4^{-}}\left(\frac{\frac{d}{dx}\left(\sqrt{16-x^{2}}\right)}{\frac{d}{dx}\left(x-4\right)}\right)$$
- step2: Find the derivative:
  $$\lim _{x\rightarrow 4^{-}}\left(\frac{-\frac{x}{\sqrt{16-x^{2}}}}{1}\right)$$
- step3: Simplify the expression:
  $$\lim _{x\rightarrow 4^{-}}\left(\frac{-x}{\sqrt{16-x^{2}}}\right)$$
- step4: Rewrite the expression:
  $$\frac{\lim _{x\rightarrow 4^{-}}\left(-x\right)}{\lim _{x\rightarrow 4^{-}}\left(\sqrt{16-x^{2}}\right)}$$
- step5: Calculate:
  $$\frac{-4}{\lim _{x\rightarrow 4^{-}}\left(\sqrt{16-x^{2}}\right)}$$
- step6: Calculate:
  $$-\infty$$
The limit of the expression $$\frac{\sqrt{16-x^{2}}}{x-4}$$ as $$x$$ approaches 4 from the left is $$-\infty$$.
The limit is $$-\infty$$.

Evaluate the limit, using L’Hôpital’s Rule if necessary.

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