Contents
Tutorial Template
by: Michael Yeh, proud Member of the Math Squad.
keyword: tutorial, limit, function, sequence
INTRODUCTION Provided here is a brief introduction to the concept of "limit," which features prominently in calculus. We first discuss the limit of a function at a point; to help motivate the definition, we first consider continuous functions. Unless otherwise mentioned, all functions here will have domain and range $ \mathbb{R} $, the real numbers. Words such as "all," "every," "each," "some," and "there are" are quite important here; read carefully!
Continuous functions
Let's consider the the following three functions along with their graphs (in blue). The red dots in each correspond to $ x=0 $, e.g. for $ f $, the red dot is the point $ (0,f(0))=(0,0) $.
- $ \displaystyle f(x)=x^3 $
- $ g(x)=\begin{cases}-x^2-\frac{1}{2} &\text{if}~x<0\\ x^2+\frac{1}{2} &\text{if}~x\geq 0\end{cases} $
- $ h(x)=\begin{cases} \sin\left(\frac{1}{x}\right) &\text{if}~x\neq 0\\ 0 &\text{if}~x=0\end{cases} $
We can see from the graphs that $ f $ is "continuous" at $ 0 $, and that $ g $ and $ h $ are "discontinuous" at 0. But, what exactly do we mean? Intuitively, $ f $ seems to be continuous at $ 0 $ because $ f(x) $ is close to $ f(0) $ whenever $ x $ is close to $ 0 $. On the other hand, $ g $ appears to be discontinuous at $ 0 $ because there are points $ x $ which are close to $ 0 $ but such that $ g(x) $ is far away from $ g(0) $. The same observation applies to $ h $.
Let's make these observations more precise. First, we will try to estimate $ f(0) $ with error at most $ 0.25 $, say. In the graph of $ f $, we have marked off a band of width $ 0.5 $ about $ f(0) $. So, any point in the band will provide a good approximation here. As a first try, we might think that if $ x $ is close enough to $ 0 $, then $ f(x) $ will be a good estimate of $ f(0) $. Indeed, we see from the graph that for any $ x $ in the interval $ (-\sqrt[3]{0.25},\sqrt[3]{0.25}) $, $ f(x) $ lies in the band (or if we wish to be more pedantic, we would say that $ (x,f(x)) $ lies in the band). So, "close enough to $ 0 $" here means in the interval $ (-\sqrt[3]{0.25},\sqrt[3]{0.25}) $; note that any point which is close enough to $ 0 $ provides a good approximation of $ f(0) $.
Can we do the same for $ g $?
TOPIC 3
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TOPIC 2
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REFERENCES
[1] "Loream Ipsum" <http://www.lipsum.com/>.
Questions and comments
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