Topics within Calculus
Aside from the intuitive definition we just saw, there is also a formal definition of a limit, known as the delta-epsilon definition. Try dragging the $\delta$ (delta) slider below, and notice what happens to the rest of the graph. But before we get to the delta-epsilon definition, let's consider what a limit really is. Any thoughts on what's going on behind the scenes?
Suppose a function $f(x)$ is undefined at $$x=0,$$ but it is continuously defined everywhere else. One such function is $$f(x)=\frac{x^{3}+2x^{2}+x}{x}.$$ However, we know that $f(x)$ is defined at $$x=\delta,$$ where $\delta$ is a non-zero value. Think of $\delta$ as a car traveling towards $0$. As $\delta$ gets closer and closer to its destination, the value of $f(\delta)$ also gets closer and closer to some other destination, which we are trying to find. So, if $\delta$ was infinitesimally small, then $f(\delta)$ would pretty much equal that destination. Hence, we can substitute $\delta$ for some infinitesimally small value, and easily see what the function evaluates to, right?
Yes, it would work, but we must be careful: For example, we know that it's okay to divide by our nonzero $\delta$, but how do we do it? How can we divide by an infinitesimally small number? In our example of $$f(x)=\frac{x^{3}+2x^{2}+x}{x},$$ it's actually very simple: $$f(x)=\frac{x^{3}+2x^{2}+x}{x} \rightarrow \frac{\delta^{3}+2\delta^{2}+\delta}{\delta} =\delta^{2}+2\delta+1.$$ Because $\delta$ is almost $0$, we can say that $$\lim_{\delta\to 0} \frac{\delta^{3}+2\delta^{2}+\delta}{\delta}=\lim_{\delta\to 0}\delta^{2}+2\delta+1=1,$$ because $$\delta^2+2\delta +1= \delta(\delta+2)+1 \rightarrow 0(0+2) +1=1.$$ As you can see above, it "looks like" $f(0)=1$. However, the value there is undefined. Would our method of finding the limit work in every situation? Why or why not? Can you think of a general rule we can use to always beee able to find the limit?
Our method above is known as the delta-epsilon definition, although we never actually met epsilon.
We say that $$\lim_{x\to c} f(x)=L$$ if, for every value $\epsilon>0$ there is a $\delta>0$ such that $$0<|x-c|<\delta$$ and $$0<|f(x)-L|<\epsilon.$$
It will likely take a while to wrap your head around that, but try to play around with the definition and some of the problems on the next page, and you'll get the hang of it. See you there!