Intuitive limits: the idea of approach

We say that the limit of $f(x)$ as $x$ approaches $a$ is $L$ , written $\lim_{x \to a} f(x) = L$ , if $f(x)$ becomes (and stays) arbitrarily close to $L$ as $x$ becomes arbitrarily close to $a$ (but not equal to $a$).

The crucial idea is "but not equal to $a$". The limit is about approach, not arrival. It does not depend on what $f(a)$ is (or whether $f(a)$ is even defined).

Two illustrations

Example 1. $f(x) = x + 2$. As $x \to 3$, $f(x) \to 5$. Easy , substitute.

Example 2. $f(x) = \dfrac{x^2 - 1}{x - 1}$ at $x = 1$. The function is undefined at $x = 1$ (denominator zero). But for $x \ne 1$, $f(x) = \dfrac{(x-1)(x+1)}{x-1} = x + 1$. So as $x \to 1$, $f(x) \to 2$. The limit is $2$ even though $f(1)$ does not exist.

One-sided limits

Sometimes we need to know whether $x$ approaches $a$ from the left ($x < a$) or the right ($x > a$).

• $\lim_{x \to a^-} f(x)$: left-hand limit.
• $\lim_{x \to a^+} f(x)$: right-hand limit.

A limit $\lim_{x \to a} f(x)$ exists iff both one-sided limits exist and are equal: $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = L \quad \Rightarrow \quad \lim_{x \to a} f(x) = L.$

When limits fail to exist

Three common failure modes:

1. Jump discontinuity. The two one-sided limits exist but differ. E.g., $f(x) = \begin{cases} 0 & x < 1 \\ 1 & x \ge 1 \end{cases}$ at $x = 1$.
2. Infinity. $f(x)$ grows without bound. E.g., $\lim_{x \to 0} \dfrac{1}{x^2} = +\infty$. (We say the limit is $\infty$, but technically the limit does not exist as a finite number.)
3. Oscillation. $f(x)$ wiggles infinitely. E.g., $\lim_{x \to 0} \sin(1/x)$ does not exist , it bounces between $-1$ and $1$.

Algebra of limits (informal)

If $\lim_{x \to a} f(x) = L$ and $\lim_{x \to a} g(x) = M$, then:

• $\lim_{x \to a} [f(x) + g(x)] = L + M$.
• $\lim_{x \to a} [f(x) - g(x)] = L - M$.
• $\lim_{x \to a} [f(x) g(x)] = L \cdot M$.
• $\lim_{x \to a} [f(x) / g(x)] = L/M$, provided $M \ne 0$.
• $\lim_{x \to a} [c f(x)] = c L$ for any constant $c$.

For elementary functions (polynomials, rational, exponentials, sine, cosine), the limit at any point in the domain is simply the value of the function there: $\lim_{x \to a} f(x) = f(a)$.

Worked examples

Example 1. $\lim_{x \to 2} (3 x + 1) = 3 \cdot 2 + 1 = 7$.

Example 2. $\lim_{x \to 0} \dfrac{x^2 + 3 x}{x} = \lim_{x \to 0} (x + 3) = 3$. (We cancelled $x$ from numerator and denominator, valid because $x \ne 0$ during the approach.)

Example 3. Find $\lim_{x \to 1^-} f(x)$ and $\lim_{x \to 1^+} f(x)$ for $f(x) = \begin{cases} 2 x & x < 1 \\ 5 & x = 1 \\ x^2 + 3 & x > 1 \end{cases}$.

Left limit: $\lim_{x \to 1^-} 2 x = 2$. Right limit: $\lim_{x \to 1^+} (x^2 + 3) = 4$. The two-sided limit does not exist.

Example 4. $\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} = \lim_{x \to 3} (x + 3) = 6$.

Example 5. Show that $\lim_{x \to 0} \dfrac{|x|}{x}$ does not exist.

For $x > 0$: $|x|/x = 1$. For $x < 0$: $|x|/x = -1$. So left limit is $-1$, right limit is $+1$ , unequal. Two-sided limit does not exist.

Try it yourself

1. $\lim_{x \to 2} (4 x - 1)$.
2. $\lim_{x \to 0} (5 - 2 x)$.
3. $\lim_{x \to 1} \dfrac{x^2 - 1}{x - 1}$.
4. $\lim_{x \to -2} \dfrac{x^2 + 5 x + 6}{x + 2}$.
5. $\lim_{x \to 4} \sqrt{x + 5}$.
6. $\lim_{x \to 0} \dfrac{x^3 - 2 x}{x}$.
7. Find $\lim_{x \to 0^-}$ and $\lim_{x \to 0^+}$ of $f(x) = \dfrac{x}{|x|}$.
8. Find $\lim_{x \to 1} f(x)$ where $f(x) = x + 1$ for $x \ne 1$ and $f(1) = 10$. (Does the value at $x = 1$ matter?)
9. $\lim_{x \to 0} \dfrac{1}{x^2}$ , what happens?
10. $\lim_{x \to 3} \dfrac{x - 3}{|x - 3|}$ , does the two-sided limit exist?
11. $\lim_{x \to 5} \dfrac{x^2 - 25}{x - 5}$.
12. If $\lim_{x \to a} f(x) = 3$ and $\lim_{x \to a} g(x) = 4$, find $\lim_{x \to a} [2 f(x) - g(x)]$.

Pitfalls / Tricks

• A limit can exist even if $f(a)$ is undefined.
• For piecewise functions, always check both one-sided limits.
• For an algebraic expression of the $0/0$ form, factor and cancel the common factor.
• Insight. A limit is about near the point, not at the point. This is the foundation of calculus.