Suppose ${\displaystyle\lim_{x\to a}}\text{ }f\left(x\right)=L$. Prove that $\underset{x\to a}{\lim }\text{ (}c\left(f\left(x\right)\right)=cL$, where $c$ is a constant.

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1. Start by recalling the definition of a limit: \( \lim_{x \to a} f(x) = L \) means that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon \).
2. To prove \( \lim_{x \to a} (c \cdot f(x)) = cL \), we need to show that for every \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |c \cdot f(x) - cL| < \epsilon \).
3. Notice that \( |c \cdot f(x) - cL| = |c| \cdot |f(x) - L| \). We can use the property of absolute values that \( |c| \cdot |f(x) - L| < \epsilon \) if \( |f(x) - L| < \frac{\epsilon}{|c|} \).
4. Since \( \lim_{x \to a} f(x) = L \), for every \( \epsilon' > 0 \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |f(x) - L| < \epsilon' \). Choose \( \epsilon' = \frac{\epsilon}{|c|} \).
5. Thus, for this choice of \( \epsilon' \), there exists a \( \delta > 0 \) such that if \( 0 < |x - a| < \delta \), then \( |c \cdot f(x) - cL| = |c| \cdot |f(x) - L| < |c| \cdot \frac{\epsilon}{|c|} = \epsilon \). This completes the proof that \( \lim_{x \to a} (c \cdot f(x)) = cL \).