The limit of something, is the maximum amount of something you can have. It can be natural or artificial. There is usually a drink drive limit, which is an artificial level of alcohol set by the state, that you are allowed to have in your system and legally drive. So if you have more alcohol than that you are known as being over the limit. Then there are natural limits, such as the limit of human hearing - dogs can hear sounds beyond that limit.
In maths limits are important because they help us get the answers to questions that we cannot ask. If that sounds odd, it is supposed to. There are some questions that we want to ask, but if we do we get nonsensical answers. As an example, lets build a function. Lets start with the function we used when talking about functions the first time:
\[x+1\]
Now, lets change the sign from addition to subtraction:
\[x-1\]
Lets divide it by itself:
\[\frac{x-1}{x-1}\]
And finally lets multiply the \(x\) on top by itself once.
\[\frac{x^2-1}{x-1}\]
Right. Lets call this \(f(x)\). What is \(f(2)\)? Well it is two times itself (four) less one (three) divided by two less one (one), so three divided by one. Or in algebra:
\[\frac{2^2-1}{2-1}\]
\[\frac{4-1}{2-1}\]
\[\frac{3}{1}\]
\[3\]
What \(f(5)\)? Well...
\[\frac{5^2-1}{5-1}\]
\[\frac{25-1}{5-1}\]
\[\frac{24}{4}\]
\[6\]
Looking at this there seems to be an obvious connection between \(x\) and \(f(x)\). The function just seems to add one to \(x\). 3=2+1, and 6=5+1. We started with \(x+1\), and it appears we have got back there! So \(f(1)\) should be two. Lets see if it is:
\[\frac{1^2-1}{1-1}\]
\[\frac{1-1}{1-1}\]
\[\frac{0}{0}\]
Hmm. Zero divided by zero. That's just a mess. Is that infinity or zero? It is meaningless. It does not work because as soon as \(x\) is one, you get zeros everywhere because you are subtracting one from one on the top and bottom. So \(f(1)\) is special because of the way the function is built.
How does the concept of a limit help us to get an answer to \(f(1)\)? Well, we could ask what \(f(0.5)\) and \(f(1.5)\) are. The answers are 1.5 and 2.5 respectively. That matches what we thought - the function ends up adding one to your starting number. It also supports our guess for \(f(1)\) before we tried it, which was two. Lets get a bit closer to one, and try again this time with \(f(0.9)\) and \(f(1.1)\). The results are 1.9 and 2.1 respectively. Again, just what we would expect. The closer we get to \(f(1)\) the closer the result gets to two. If we try with something silly like \(f(0.99999)\) we get 1.99999. That pattern is going to repeat for as many nines as we add onto the end of the number we plug in.
The way we answer this question is to say that as the number we plug into the function gets closer and closer to one, the result of the function gets closer and closer to two, but because it breaks when we put in exactly one, we never get to exactly two. That's what a limit is. In mathematical script it looks like this:
\[\lim_{x \rightarrow 1} \frac{x^2-1}{x-1}=2\]
Spoken out loud that says the limit of the function (blah blah describe the function blah blah) as ecks approaches one is two.
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