6. $\\lim_{x \\to 0} \\frac{\\frac{1}{x + 2} - \\frac{1}{2}}{x}$

Explanation:

Step1: Combine the fractions in the numerator

First, find a common denominator for (\frac{1}{x + 2}-\frac{1}{2}). The common denominator is (2(x + 2)). [ \begin{align*} \frac{1}{x+2}-\frac{1}{2}&=\frac{2-(x + 2)}{2(x + 2)}\ &=\frac{2-x - 2}{2(x + 2)}\ &=\frac{-x}{2(x + 2)} \end{align*} ] So the original limit (\lim_{x\rightarrow0}\frac{\frac{1}{x + 2}-\frac{1}{2}}{x}) becomes (\lim_{x\rightarrow0}\frac{\frac{-x}{2(x + 2)}}{x}).

Step2: Simplify the complex - fraction

When we have (\frac{\frac{-x}{2(x + 2)}}{x}), this is equivalent to (\frac{-x}{2(x + 2)}\cdot\frac{1}{x}) (since dividing by (x) is the same as multiplying by (\frac{1}{x})). Cancel out the non - zero (x) terms ((x\neq0) as we are taking the limit as (x\rightarrow0), not evaluating at (x = 0)). [ \frac{-x}{2(x + 2)}\cdot\frac{1}{x}=\frac{-1}{2(x + 2)} ]

Step3: Evaluate the limit

Now we find (\lim_{x\rightarrow0}\frac{-1}{2(x + 2)}). Substitute (x = 0) into the function (\frac{-1}{2(x + 2)}). [ \lim_{x\rightarrow0}\frac{-1}{2(x + 2)}=\frac{-1}{2(0 + 2)} ]

Answer:

(-\frac{1}{4})