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I am given that:

$f(x)=cx$, where $c$ is a constant

Show that $\lim_{x→x_0}f(x)=f(x_0)$

I'm not really sure how to start this, any help would be greatly appreciated!

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    Start with the definition of limit. In your question, write down the definition of limit you are using. – 2017-01-18
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    @Czar You see the check mark at the left of my answer? Click it if you think my answer is correct. Thank you – 2017-01-18

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Let $\epsilon>0$. Take $\delta=\frac{\epsilon}{|c|+1}$. If $0<|x-x_0|<\delta$ then $$|f(x)-f(x_0)|=|cx-cx_0|=|c||x-x_0|<|c|\cdot\delta=|c|\cdot\frac{\epsilon}{|c|+1}<\epsilon.$$

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    Awesome! How do you determine what delta is? – 2017-01-18
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    We have $|f(x)-f(x_0)|=|c||x-x_0|$ and we want it to be less than $\epsilon$. Now, we know that we have to assume that $0<|x-x_0|<\delta$. With this, we get $$|f(x)-f(x_0)|<|c|\delta.$$ So we need to look for $\delta$ for which $|c|\delta<\epsilon$. I took $\delta=\frac{\epsilon}{|c|+1}$, because $c$ might be 0. – 2017-01-18
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    I'm starting to understand it quite better now. What if c is stated that it cannot equal 0? – 2017-01-18
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    Actually, the proof covers either $c=0$ or $c\neq 0$. – 2017-01-18
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    Note that $\frac{|c|}{|c|+1}<1$. – 2017-01-18
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    Got it, thank you very much! – 2017-01-18