Question

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

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 \(2(x + 2)\).

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So the original limit \(\lim_{x
ightarrow0}\frac{\frac{1}{x + 2}-\frac{1}{2}}{x}\) becomes \(\lim_{x
ightarrow0}\frac{\frac{-x}{2(x + 2)}}{x}\).

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 (\(x
eq0\) as we are taking the limit as \(x
ightarrow0\), 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 \(\lim_{x
ightarrow0}\frac{-1}{2(x + 2)}\). Substitute \(x = 0\) into the function \(\frac{-1}{2(x + 2)}\).

$$ \lim_{x ightarrow0}\frac{-1}{2(x + 2)}=\frac{-1}{2(0 + 2)} $$

Answer:

\(-\frac{1}{4}\)