Prove that \begin{equation}\lim_{x\to a}\,(m\,x+b)=m\,a+b \end{equation} whem $m$ is not zero.

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Lesson Parent: 
Function Limits III: The Gory Details
Problem: 

Prove that \begin{equation}\lim_{x\to a}\,(m\,x+b)=m\,a+b \end{equation} whem $m$ is not zero.

Answer: 

It is true that \begin{equation}\lim_{x\to a}\,(m\,x+b)=m\,a+b \end{equation} whem $m$ is not zero.

SAMPLE SOLUTION

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Solution: 

Here are the general set of steps we need to take to do our proof. Here are the steps for this problem:

Step Description
1 Because

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and <Sign in to see all the formulas> we will make the guess that <Sign in to see all the formulas>. This is why we make the restriction that <Sign in to see all the formulas>.
2 Say <Sign in to see all the formulas>.
3 <Sign in to see all the formulas>
4 Now assume that we are at some value for $x$ such that <Sign in to see all the formulas>.
5 Now plug the function <Sign in to see all the formulas> into <Sign in to see all the formulas> so that we have <Sign in to see all the formulas>.
6 Let's do some algebra:

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7 Since $C$ is arbitrary we have a <Sign in to see all the formulas> for every <Sign in to see all the formulas> such that when <Sign in to see all the formulas> then <Sign in to see all the formulas>.

Therefore, we have proven that the statement

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is true.

But what about the case when $m=0$? In that case

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and we are trying to find the limit of a constant number. Here is the proof.

Now because

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a straight line is a continuous function.

Finally, because we never specified what the value of $a$ was, it is an arbitrary number. Therefore, a straight line is continuous on the entire x-axis.