Example 1: a polynomial function

• Find lim_{x → 2} (x² + 1).
• Solution This is a polynomial, so we can substitute directly: 2² + 1 = 5. The limit is 5.

Example 2: a function not defined at the point

• Find lim_{x → 1} (x² − 1) / (x − 1).
• Solution At x = 1 both the numerator and the denominator are zero, so direct substitution is not possible. However, x² − 1 = (x − 1)(x + 1), so the fraction equals x + 1 for every x ≠ 1. The limit is 1 + 1 = 2.
• The table confirms this:

1. The values approach 2 from both sides. The function is undefined at 1
2. but the limit exists.

Example 3: different limits from the two sides

• Let f(x) = |x| / x for x ≠ 0. Find the limit at 0, if it exists.
• Solution From the right, when x > 0 we have |x| = x, so f(x) = 1. Therefore lim_{x → 0⁺} f(x) = 1. From the left, when x < 0 we have |x| = −x, so f(x) = −1. Therefore lim_{x → 0⁻} f(x) = −1.
• The two one-sided limits are different, so the limit at 0 does not exist.

Example 4: the limit is different from the function value

• Let f(x) = 3x + 1 when x < 2, f(2) = 5, and f(x) = 9 − x when x > 2. Find lim_{x → 2} f(x) and compare it with f(2).

• Solution From the left: 3 × 2 + 1 = 7. From the right: 9 − 2 = 7. The two one-sided limits are equal, so lim_{x → 2} f(x) = 7.
• But the value of the function at 2 is f(2) = 5.
• The limit (7) is not equal to the function value (5), which shows the difference between the two ideas.

Example 5: a limit that does not exist because the values grow without bound

• Find lim_{x → 3} 1 / (x − 3), if it exists.
• Solution When x approaches 3 from the right, the denominator is a small positive number, so the value becomes very large and positive. From the left, the denominator is a small negative number, so the value becomes very large and negative. The one-sided limits are different and infinite, so the limit does not exist.