Indeterminate Limits I have been studying independently through various online courses and I still have trouble understanding what to do with certain limits. I am hoping for some guidance on the following two problems to help me solve them (I do not need to answer as much as help understanding where I am going).
$$ \lim_{x\to 0}  \frac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x} $$
$$ \lim_{x\to 0} \frac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x} $$
I can tell that these are both $ \frac{0}{0} $ indeterminate equations and I can simplify them in a number of ways but I cannot seem to get the 0 out of the bottom of the equation. Any help pointing me in the right direction would be greatly appreciated!

I would assume that you typed in the expression for the limit correctly.
For the problem you provided, there is a short way to determine the limit. Given that the limit is:
$$\lim_{x\to 0}  \frac{\sqrt{16 + 4x} - \sqrt{16 + 4x}}{x}$$
The numerator expression becomes $0$.  Then, we have:
$$\lim_{x\to 0} \frac{0}{x} = \lim_{x \to 0} 0 = 0$$
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For the second problem, we can rewrite the expression as:
$$\lim_{x \to 0} \dfrac{\frac{1}{(x + 6)^2} - \frac{1}{36}}{x} \cdot \dfrac{36(x + 6)^2}{36(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{36 - (x + 6)^2}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{36 - x^2 - 12x - 36}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{-x(x + 12)}{36x(x + 6)^2}$$
$$= \lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2}$$
Thus,
$$\lim_{x \to 0} \dfrac{-(x + 12)}{36(x + 6)^2} = \dfrac{-12}{36(6)^2} = -\dfrac{1}{108}$$
Verified by Wolfram for:
First problem
Second problem
For the problem you edited
The given limit is:
$$\lim_{x \to 0} \dfrac{\sqrt{16 + 4x} - \sqrt{16 - 4x}}{x}$$
Multiply the top and bottom by the conjugate of the numerator expression, which is $\sqrt{16 + 4x} + \sqrt{16 - 4x}$.  This gives:
$$\lim_{x \to 0} \dfrac{16 + 4x - (16 - 4x)}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$
$$= \lim_{x \to 0} \dfrac{8x}{x(\sqrt{16 + 4x} + \sqrt{16 - 4x})}$$
$$= \lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$
Thus,
$$\lim_{x \to 0} \dfrac{8}{\sqrt{16 + 4x} + \sqrt{16 - 4x}}$$
$$= \dfrac{8}{\sqrt{16} + \sqrt{16}}$$
$$= \dfrac{8}{4 + 4}$$
$$= 1$$
Verified by WolframAlpha.
