In this post, we discuss the concept of the limit of a function and review the limit laws. The limit of a function is to investigate the behavior of a function near a point. This is even though the value of the function at that point is not definable or unknown. What we say about single-variable functions can easily be expressed for other kinds of functions discussed in multivariable calculus.

#### An informal definition of the limit of a function

Let \(f\) be a function, definable near a point \(c\) except possibly at \(c\) itself. If \(f(x)\) is close enough to \(L\) for all \(x\) sufficiently close to \(c\), then we say the function \(f\) approaches to the limit \(L\) whenever \(x\) approaches to \(x\) and we denote this by $$\lim_{x \to c} f(x) = L.$$

For example, in high school mathematics, we see that the function \(f\) defined by $$f(x) = \frac{\sin x}{x}$$ approaches to 1, whenever \(x\) approaches to 0 though the fraction \(\frac{\sin x}{x}\) is not definable at 0. We denote this by the following limit formula: $$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$

#### The laws of the limit of a function

In the following, we list the basic laws of the limit of a function:

**Result**. Let \(f\) and \(g\) be functions with $$\lim_{x \to c} f(x) = P$$ and $$\lim_{x \to c} g(x) = Q.$$ Then, the following statements hold:

- \(\lim_{x \to c} (f+g)(x) = P + Q.\)
- \(\lim_{x \to c} (f-g)(x) = P – Q.\)
- \(\lim_{x \to c} (fg)(x) = PQ.\)
- \(\lim_{x \to c} (kf)(x) = kP,\) if \(k\) is a real constant.
- \(\lim_{x \to c} (f/g)(x) = P/Q,\) if \(Q \neq 0\).
- \(\lim_{x \to c} (f^n)(x) = P^n.\)
- \(\lim_{x \to c} (\sqrt[n]{f})(x) = \sqrt[n]{P}.\) Note that if \(n\) is an even number, we suppose that \(f\) has non-negative values near \(c\).

**Exercise**. Find the limit of the following functions at the given points:

(a) \(f(x) = x^3 + \sqrt{x+1} + \displaystyle \frac{\tan x}{x}\) at \(x = 0\).

(b) \(g(x) = \displaystyle \frac{x^n – 1}{x-1}\) at \(x = 1\).

**Solution**. (a) It is clear that $$\lim_{x \to 0} x^3 = 0^3 = 0,$$ and $$\lim_{x \to 0} \sqrt{x+1} = \sqrt{0+1} = 1.$$ On the other hand, $$\lim_{x \to 0} \frac{\tan x}{x} = $$ $$\lim_{x \to 0} \frac{1}{\cos x}\frac{\sin x}{x} = $$ $$\frac{1}{\cos 0} \cdot 1 = 1.$$ Therefore, $$\lim_{x \to 0} f(x) = $$ $$ \lim_{x \to 0} \left(x^3 + \sqrt{x+1} + \frac{\tan x}{x}\right) = $$ $$ 2.$$

(b) Since $$\frac{x^n – 1}{x-1} = $$ $$x^{n-1} + \dots + x + 1,$$ we see that $$\lim_{x \to 1} \frac{x^n – 1}{x-1} = $$ $$ \sum_{i=0}^{n-1} 1^i = n.$$

#### An important rule for the limit of a function

**The sandwich rule**: Let \(f\), \(g\), and \(h\) be functions with $$f(x) \leq g(x) \leq h(x)$$ for all \(x\) near a point \(c\). Also, suppose that $$\lim_{x \to c} f(x) = \lim_{x \to c} h(x) = L.$$ Then \(\lim_{x \to c} g(x) = L\).

**Remark**. Some references refer to “the sandwich rule” as “the squeeze rule” or ” the pinching rule”.

**Exercise**. Let \(g\) be a function satisfying the following inequality: $$\sqrt[3]{27-2x^2} \leq g(x) \leq \sqrt[3]{27-x^2}.$$ Find \(\lim_{x \to 0} g(x).\)

**Solution**. It is clear that $$\lim_{x \to 0} \sqrt[3]{27-2x^2} = \sqrt[3]{27} = 3$$ and $$\lim_{x \to 0} \sqrt[3]{27-x^2} = \sqrt[3]{27} = 3.$$ Therefore, by the sandwich rule, \(\lim_{x \to 0} g(x) = 3.\)

### The precise definition of the limit of a function

Let \(f\) be a function and \(c\) be a real number such that an open neighborhood of \(c\), except possibly the point \(c\) itself, is in the domain \(D_f\) of the function \(f\), i.e. there is a positive real number \(r\) with $$ (c-r, c+r) \setminus \{c\} \subseteq D_f.$$ We say the limit of \(f\) is \(L\) as \(x\) approaches to \(c\) and write $$\lim_{x \to c} f(x) = L$$ if for every positive real number \(\epsilon\), there is a positive real number \(\delta\) (most probably depending on \(\epsilon\)) such that the assumption $$ 0 < \vert x – c \vert < \delta $$ implies $$ \vert f(x) – L \vert < \epsilon.$$

The above definition is sometimes referred to as the epsilon-delta definition of the limit.

**Exercise**. Using the epsilon-delta definition, show that $$\lim_{x \to 1} (2x+1) = 3.$$

**Solution**. Given the positive real number \(\epsilon\), we need to prove that $$ \vert (2x+1) – 3 \vert < \epsilon,$$ considering the assumption $$ 0 < \vert x – 1 \vert < \delta $$ for some suitable positive real number \(\delta\). Observe that $$ \vert (2x+1) – 3 \vert = 2 \vert x – 1 \vert.$$ So, if we take \(\delta = \epsilon / 2\), the assumption $$ 0 < \vert x – 1 \vert < \delta = \epsilon / 2 $$ implies that $$ 2 \vert x – 1 \vert < \epsilon. $$

This is equivalent to say that $$ \vert (2x+1) – 3 \vert < \epsilon.$$

**Exercise**. Let \(\lim_{x \to c} f(x) = L\). Using delta-epsilon definition, prove that $$\lim_{x \to c} (f(x))^2 = L^2.$$

#### The precise definition of the limit of a function at infinity

Let \(f: (a,+\infty) \rightarrow \mathbb{R}\) be a function and \(L\) be a real number. We say the limit of \(f\) is \(L\) as \(x\) approaches to \(+\infty\) and write $$\lim_{x \to +\infty} f(x) = L$$ if for every positive real number \(\epsilon\), there is a positive real number \(M\) (most probably depending on \(\epsilon\)) such that the assumption $$ x > M $$ implies $$ \vert f(x) – L \vert < \epsilon.$$

Let \(f: (-\infty, a) \rightarrow \mathbb{R}\) be a function and \(L\) be a real number. We say the limit of \(f\) is \(L\) as \(x\) approaches to \(-\infty\) and write $$\lim_{x \to -\infty} f(x) = L$$ if for every positive real number \(\epsilon\), there is a positive real number \(M\) (most probably depending on \(\epsilon\)) such that the assumption $$ x < -M $$ implies $$ \vert f(x) – L \vert < \epsilon.$$

Finally, let \(a \leq b\), \(f: (-\infty, a) \cup (b,+\infty) \rightarrow \mathbb{R}\) be a function, and \(L\) a real number. We say the limit of \(f\) is \(L\) as \(x\) approaches to \(\infty\) and write $$\lim_{x \to \infty} f(x) = L$$ if $$\lim_{x \to +\infty} f(x) = \lim_{x \to -\infty} f(x) = L.$$

#### Continuous functions

By definition, a function \(f\) is continuous at \(c\) if the limit of the function \(f\) at \(c\) exists and $$\lim_{x \to c} f(x) = f(c).$$

**Example**. Any single-variable polynomial function is continuous at each real point.

**Exercise**. Prove that if \(f\) is continuous and 1-to-1 on at each element of an open interval, then \(f^{-1}\) is also continuous.