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.

Verified step by step guidance

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