# Definition of the Limit of a Function

We've already talked about the limit of a sequence, and, since a sequence is a particular case of a function, there will be a similarity between the sequence and the function. We are also going to extend some concepts.

**Example 1.** Let's investigate the behavior of $$${y}={{x}}^{{2}}$$$ near the point $$${x}={2}$$$.

To do this, let's take the points sufficiently close to $$$2$$$ and evaluate the corresponding y-values.

$$${x}$$$ | 1.9 | 1.99 | 1.999 | 2.0001 | 2.001 | 2.01 |

$$${y}={{x}}^{{2}}$$$ | 3.61 | 3.9601 | 3.996001 | 4.00040001 | 4.004001 | 4.0401 |

We see that as $$${x}$$$ approaches $$$2$$$, $$${f{{\left({x}\right)}}}$$$ approaches $$$4$$$, no matter whether we take the points greater than $$$2$$$ or less than $$$2$$$. We write this fact as $$$\lim_{{{x}\to{2}}}{{x}}^{{2}}={4}$$$.

**Definition.** We write that $$$\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L}$$$ and say, "the limit of $$$f(x)$$$, as $$$x$$$ approaches $$$a$$$, equals $$$L$$$" if for any $$$\epsilon>{0}$$$ there exists $$$\delta>{0}$$$ such that $$${\left|{f{{\left({x}\right)}}}-{L}\right|}<\epsilon$$$ when $$${\left|{x}-{a}\right|}<\delta$$$.

To put it simply, this means that $$$\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}={L}$$$ if we can make $$${f{{\left({x}\right)}}}$$$ as close to $$${L}$$$ as we like by taking an $$${x}$$$ sufficiently close to $$${a}$$$.

We will also use the following notation: $$${f{{\left({x}\right)}}}\to{L}$$$ as $$${x}\to{a}$$$.

From the definition, it follows that we only care about the behavior of the function at the points near $$${a}$$$, not at $$${a}$$$ itself. This means that $$${f{{\left({x}\right)}}}$$$ can even be not defined at $$${a}$$$ but still have $$$\lim_{{{x}\to{a}}}{f{{\left({x}\right)}}}$$$.

For example, consider the function $$${f{{\left({x}\right)}}}={{x}}^{{2}}$$$ and $$${g{{\left({x}\right)}}}={\left\{\begin{array}{c}{{x}}^{{2}}{\quad\text{if}\quad}{x}\ne{2}\\\text{undefined}{\quad\text{if}\quad}{x}={2}\\ \end{array}\right.}$$$.

Note that $$$\lim_{{{x}\to{2}}}{f{{\left({x}\right)}}}=\lim_{{{x}\to{2}}}{g{{\left({x}\right)}}}={4}$$$ even despite the fact that $$${g{{\left({x}\right)}}}$$$ is not defined at $$$2$$$. That's because we don't care about the function at $$$2$$$, we are only interested in the bevavior near $$$2$$$.

Note that $$${\left|{x}-{a}\right|}<\delta$$$ is equivalent to the double inequality $$$-\delta<{x}-{a}<\delta$$$ or $$${a}-\delta<{x}<{a}+\delta$$$.

Let's do a couple of examples trying to guess the limit. Note that this approach is not correct but it allows to understand better the limits.

**Example 2.** Calculate $$$\lim_{{{x}\to{1}}}{\left({x}+{1}\right)}$$$.

Let's see what value $$${\left({x}+{1}\right)}$$$ approaches as $$${x}$$$ approaches $$$1$$$.

$$$x$$$ | 0.9 | 0.99 | 0.999 | 1.0001 | 1.001 | 1.01 |

$$${y}={x}+{1}$$$ | 1.9 | 1.99 | 1.999 | 2.0001 | 2.001 | 2.01 |

So, as $$${x}$$$ approaches $$$1$$$, $$${f{{\left({x}\right)}}}$$$ approaches $$$2$$$; thus, $$$\lim_{{{x}\to{1}}}{\left({x}+{1}\right)}={2}$$$.

One more quick example.

**Example 3.** Calculate $$$\lim_{{{x}\to{1}}}\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$$$.

Let's see what value $$$\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$$$ approaches when $$${x}$$$ approaches $$$1$$$.

$$$x$$$ | 0.9 | 0.99 | 0.999 | 1.0001 | 1.001 | 1.01 |

$$${y}=\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}$$$ | $$$\frac{{{0.9}-{1}}}{{{{0.9}}^{{2}}-{1}}}={0.5263}$$$ | 0.5025 | 0.5003 | 0.499975 | 0.49975 | 0.4975 |

So, as $$$x$$$ approaches $$$1$$$, $$${f{{\left({x}\right)}}}$$$ approaches $$$0.5$$$; thus, $$$\lim_{{{x}\to{1}}}\frac{{{x}-{1}}}{{{{x}}^{{2}}-{1}}}=\frac{{1}}{{2}}$$$.

And another useful example.

**Example 4.** Calculate $$$\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}$$$.

Let's see where $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}$$$ goes when $$${x}$$$ approaches $$$0$$$.

$$${x}$$$ | $$${f{{\left({x}\right)}}}$$$ |

1 | $$${\cos{{\left(\frac{\pi}{{1}}\right)}}}=-{1}$$$ |

0.1 | $$${\sin{{\left(\frac{\pi}{{0.1}}\right)}}}=-{1}$$$ |

0.001 | $$${\sin{{\left(\frac{\pi}{{0.001}}\right)}}}=-{1}$$$ |

0.0001 | $$${\sin{{\left(\frac{\pi}{{0.0001}}\right)}}}=-{1}$$$ |

It seems that $$$\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$$$, but this is the WRONG answer! The reason is that we took only the values at which $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$$$.

Note that $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}=-{1}$$$ when $$${x}=\frac{{1}}{{n}}$$$ for any integer $$${n}$$$.

But also $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}={0}$$$ for infinitely many values of $$${x}$$$ that approach $$$0$$$. In fact, as $$${x}$$$ approaches $$$0$$$, the cosine takes any value from the interval $$${\left[-{1},{1}\right]}$$$. In other words, $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}$$$ oscillates infinitely many times as $$${x}$$$ approaches $$$0$$$.

Since $$${\cos{{\left(\frac{\pi}{{x}}\right)}}}$$$ doesn't approach any fixed value, $$$\lim_{{{x}\to{0}}}{\cos{{\left(\frac{\pi}{{x}}\right)}}}$$$ doesn't exist.

This example shows that we can guess the wrong value if we take the inappropriate values of $$${x}$$$. It is also difficult to understand when to stop calculating the values. However, we will give the correct methods for calculating the limits.

Let's finalize our work with one last example.

**Example 5.** Calculate $$$\lim_{{{x}\to{0}}}\frac{{1}}{{{x}}^{{2}}}$$$.

As $$$x$$$ approaches $$$0$$$, $$${{x}}^{{2}}$$$ also approaches $$$0$$$; so, $$$\frac{{1}}{{{x}}^{{2}}}$$$ becomes very large (for example, when $$${x}={0.001}$$$, we have that $$$\frac{{1}}{{{x}}^{{2}}}={1000000}$$$). Thus, the value of $$${f{{\left({x}\right)}}}$$$ approaches infinity without a bound; so, $$$\lim_{{{x}\to{0}}}\frac{{1}}{{{x}}^{{2}}}=\infty$$$.