- Studying
*limits*means that we investigate the values of \(f(x)\) as \(x\) approaches a given number \(a\)- Let's do this in case \(f(x)=x^2-x+2\) and \(a=2\).
- Click here for a table of values and a graph
- You can see that the closer \(x\) gets to \(2\), the closer \(f(x)\) gets to \(4\).
- It appears that we could make \(f(x)\) get arbitrarily close to \(4\) by making \(x\) get arbitrarily close to \(2\).
- We express this by saying that
*the limit of the function \(f(x)=x^2-x+2\) as \(x\) approaches 2 is 4*. - The notation for this is \(\lim_{x\to a}f(x)=4\).

*Intuitive definition of a limit.*- Suppose that a function \(f(x)\) is defined near a number \(a\). This means that there exists an open interval \((\alpha,\beta)\) around \(a\) such that \(f(x)\) is defined whenever \(\alpha<x<\beta\), except possibly \(a\).
- Then we write \(\lim_{x\to a}f(x)=L\) and say that
*the limit of \(f(x)\) as \(x\) approaches \(a\) is \(L\)*, if we can make sure that \(f(x)\) is arbitrarily close to \(L\) by making \(x\) close enough to \(a\), but not equal to \(a\). - For example, on the table and the graph, you can see that if \(1.999<x<2.001\), then \(3.997<f(x)<4.003\).
- An alternative notation for \(\lim_{x\to a}f(x)=L\) is \(f(x)\to L\text{ as }x\to a\).

- Consider the following list of values for \(f(x)=\frac{x-1}{x^2-1}\).

x | f(x) | x | f(x) |
---|---|---|---|

0.9 | 0.526315789474 | 1.1 | 0.47619047619 |

0.999 | 0.500250125063 | 1.001 | 0.499750124938 |

0.99999 | 0.500002500013 | 1.00001 | 0.499997500012 |

0.9999999 | 0.50000002498 | 1.0000001 | 0.49999997502 |

- Our guess is that \(\lim_{x\to1}\frac{x-1}{x^2-1}=\frac12\), and that is indeed the case.
- Click here for a graph.
- Note that by definition the limit \(\lim_{x\to a}(f(x))\) not depend on whether \(f(x)\) is defined at \(x=a\), or what \(f(x=a)\) is. For example, taking \[ g(x)=\begin{cases} \frac{x-1}{x^2-1} & x\ne1 \\ 2 & x=1, \end{cases} \] we still get \(\lim_{x\to1}g(x)=\frac12\).

- Consider the following list of values for \(f(t)=\frac{\sqrt{t^2+9}-3}{t^2}\).

t | f(t) | t | f(t) |
---|---|---|---|

-0.1 | 0.1666203960726864 | 0.1 | 0.1666203960726864 |

-0.001 | 0.1666666620547819 | 0.001 | 0.1666666620547819 |

-0.0000001 | 0.1665334536937735 | 0.0000001 | 0.1665334536937735 |

-0.00000000001 | 0 | 0.00000000001 | 0 |

- Up until \(x=\pm0.000001=\pm10^{-5}\), it looks like \(\lim_{t\to0}\frac{\sqrt{t^2+9}-3}{t^2}=1.6666\ldots=\frac16\), but then the numbers drop to 0.
- Click here for the corresponding plots, where you can see a similar behaviour.
- As we will be able to prove, it is indeed the case that \(\frac{\sqrt{t^2+9}-3}{t^2}\to\frac16\) as \(x\to\frac16\).
- The zeroes you get for \(x\) very close to 0 come from the limitations of numerical calculations.
- Every calculator or computer program performs numerical calculations with numbers rounded up to a fixed number.
- For example, in case of SAGE, the program I'm using, this number format is that of
*double-precision floating points*: numbers are stored in 8 bytes. - The point is that because of this rounding procedure, when the nominator \(\sqrt{t^2+9}-3\) gets really close to zero, it is treated as being 0, before getting divided by \(t^2\).
- See more in the pdf Lies my calculator and computer told me on the textbook's webpage.

- Consider the following list of values for \(f(x)=\frac{\sin x}{x}\).

x | f(x) | x | f(x) |
---|---|---|---|

-0.1 | 0.9983341664682815 | 0.1 | 0.9983341664682815 |

-0.001 | 0.9999998333333416 | 0.001 | 0.9999998333333416 |

-0.00001 | 0.9999999999833333 | 0.00001 | 0.9999999999833333 |

-0.0000001 | 0.9999999999999983 | 0.0000001 | 0.9999999999999983 |

- Our guess is that \(\lim_{x\to0}\frac{\sin(x)}{x}=1\), which is indeed the case: we'll see how to prove this in Chapter 3.

- Consider the following list of values for \(f(x)=\sin\frac{\pi}{x}\).

x | f(x) | x | f(x) |
---|---|---|---|

-0.1 | 0.0000000000000001 | 0.1 | -0.0000000000000001 |

-0.001 | -0.0000000000000196 | 0.001 | 0.0000000000000196 |

-0.00001 | 0.0000000000012188 | 0.00001 | -0.0000000000012188 |

- We're getting numbers really close to 0, but they don't seem to get closer to 0 when \(x\) does.
- That itself is a result of numerical approximation: we can calculate these values precisely \[ f(\pm0.1)=\sin(\pm10\pi)=0,\,f(\pm0.001)=\sin(\pm1000\pi)=0,\,f(\pm0.00001)=\sin(100000\pi)=0. \]
- This could lead you to think that the limit is zero.
- But then, you can try another set of numbers getting arbitrarily close 0 zero, which give a different result \[ f(\frac{1}{10.5})=\sin(10.5\pi)=1,\,f(\frac{1}{1000.5})=\sin(1000.5\pi)=1,\,f(\frac{1}{100000.5})=\sin(100000.5\pi)=1. \]
- That is, it is not true that \(\lim_{x\to}\sin\frac{\pi}{x}=0\), because we can find numbers \(x\) arbirarily close to 0, for which \(f(x)=1\), and vice versa.
- This means that \(\lim_{x\to0}\sin\frac{\pi}{x}\) does not exist.
- Click here for a graph.

- What could be said about \(f(x)=\sqrt x\) as \(x\) approaches 0?
- The issue is that the function \(f(x)\) is defined on \([0,\infty)\), so you can only approach \(0\)
*from the right*. - It looks like we can make sure that \(f(x)\) is arbitrarily close to \(0\) if \(x\) is sufficiently close to \(a\) with \(x>a\):

- The issue is that the function \(f(x)\) is defined on \([0,\infty)\), so you can only approach \(0\)

x | f(x) |
---|---|

0.1 | 0.3162277660168379 |

0.001 | 0.0316227766016838 |

0.00001 | 0.0031622776601684 |

- This is an example of a
*one-sided limit*. Click here to see a graph.*Definition of right-hand limit.*Suppose that \(f(x)\) is defined on an open interval having \(a\) as a left endpoint. We say that*the right-hand limit of \(f(x)\) as \(x\) approaches \(a\)*(or the *limit of \(f(x)\) as \(x\) approaches \(a\) from the right) is \(L\), and we write \[ \lim_{x\to a^+}f(x)=L, \] if we can make sure that \(f(x)\) is arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) with \(x>a\).- Similarly, we can define
*left-hand limits*. The notation for those is \[ \lim_{x\to a^-}f(x)=L. \]

- Consider the following list of values for \(f(x)=\frac{1}{x^2}\).

x | f(x) | x | f(x) |
---|---|---|---|

-0.1 | 100 | 0.1 | 100 |

-0.001 | 1000000 | 0.001 | 1000000 |

-0.00001 | 10000000000 | 0.00001 | 10000000000 |

- You can see that as \(x\) approaches \(0\), the value \(f(x)\) gets arbitrarily large.
- Click here for a graph
- We denote this behaviour by \(\lim_{x\to0}\frac{1}{x^2}=\infty\).
*Intuitive definition of an infinite limit*. Let \(f(x)\) be a function defined around the number \(a\), except possibly \(a\) itself. Then we write \[ \lim_{x\to a}f(x)=\infty, \] if we can make sure that the value \(f(x)\) gets arbitrarily large, by taking \(x\) sufficiently close to \(a\).- Note that \(\infty\) is not a number, the limit does not exist. But this nonexistence happens in a particular way, which is indicated by the notation \(\lim_{x\to a}f(x)=\infty\).
- Similarly, if we can make sure that the value \(f(x)\) gets arbitrarily small by taking \(x\) sufficiently close to \(a\), we write \(\lim_{x\to a}f(x)=-\infty\). For a cheap example, consider \(f(x)=-\frac{1}{x^2}\).

- Consider the following values for \(f(x)=\tan(x)\).

x | f(x) | x | f(x) |
---|---|---|---|

\(\pi/2-0.1\) | 9.9666444232592362 | \(\pi/2+0.1\) | -9.9666444232592379 |

\(\pi/2-0.001\) | 999.999666666637836 | \(\pi/2+0.001\) | -999.9996666666493184 |

\(\pi/2-0.00001\) | 99999.9999966482282616 | \(\pi/2+0.00001\) | -99999.9999967626499711 |

- It looks like \(\lim_{x\to(\pi/2)^-}\tan(x)=\infty\) and \(\lim_{x\to(\pi/2)^+}\tan(x)=-\infty\).
- Click here for the graph.
- You can see that as you approach \(\frac{\pi}{2}\) with \(x\), the tangent line to the graph \(y=\tan(x)\) gets closer and closer to the vertical line \(x=\frac{\pi}{2}\).
- In this case, we say that the line \(x=\frac{\pi}{2}\) is a
*vertical asymptote*of \(\tan(x)\). *Definition.*We say that the line \(x=a\) is a*vertical asymptote*of the curve \(y=f(x)\), if at least one of the following statements is true. \[\begin{gather*} \lim_{x\to a}f(x)=\infty\quad\lim_{x\to a^-}f(x)=\infty\quad\lim_{x\to a^+}f(x)=\infty\\ \lim_{x\to a}f(x)=-\infty\quad\lim_{x\to a^-}f(x)=-\infty\quad\lim_{x\to a^+}f(x)=-\infty. \end{gather*}\]- Exercises. 2.2: 4, 6, 8, 16, 20, 32, 52