Calculus Limits: 12 Common Mistakes and How to Fix Them

Short answer: most lost marks on limits come from three habits: plugging in and stopping at 0/0, ignoring one-sided behaviour, and cancelling or dividing incorrectly. Always try direct substitution first; if you get 0/0 or ∞/∞, that is a signal to simplify, not an answer.

The basic routine

1. Substitute the value. A real number means you are done.
2. Got k/0 with k ≠ 0? The limit is infinite or does not exist — check both sides.
3. Got 0/0? Factor, rationalise, or use a known limit (or L'Hôpital's rule).
4. Limit at infinity? Divide by the highest power in the denominator.

The 12 mistakes

1. Writing "0/0" as the answer

0/0 is indeterminate: it tells you nothing yet. Fix: simplify. Example: lim_x→2 (x² − 4)/(x − 2) = lim (x − 2)(x + 2)/(x − 2) = lim (x + 2) = 4.

2. Saying k/0 = 0

lim_x→0 1/x² is +∞, not 0. A nonzero number over something tiny is huge.

3. Ignoring one-sided limits

lim_x→0 1/x does not exist: from the right it goes to +∞, from the left to −∞. Always check both sides when the denominator changes sign.

4. Thinking the limit equals f(a)

The limit describes what happens near a, not at a. A function with a hole at x = 3 can still have a limit there.

5. Cancelling terms instead of factors

(x² + x)/x simplifies to x + 1 because x is a common factor. You cannot "cancel" the x in (x + 1)/x.

6. Forgetting the conjugate trick

lim_x→0 (√(x + 9) − 3)/x. Multiply by (√(x + 9) + 3)/(√(x + 9) + 3): numerator becomes x, so the limit is 1/(√9 + 3) = 1/6.

7. Using degrees with sin x / x

lim_x→0 sin x / x = 1 only in radians. Variant: lim sin(5x)/x = 5 · lim sin(5x)/(5x) = 5.

8. Wrong leading terms at infinity

lim_x→∞ (3x² + 1)/(5x² − x) = 3/5. Divide everything by x². Same degree → ratio of leading coefficients; bottom higher → 0; top higher → ±∞.

9. ∞ − ∞ = 0

lim_x→∞ (√(x² + x) − x) is not 0. Using the conjugate gives x/(√(x² + x) + x) → 1/2.

10. Losing the sign with √(x²) at −∞

√(x²) = |x|, which equals −x when x is negative. lim_x→−∞ √(x² + 1)/x = −1.

11. Applying L'Hôpital when it isn't 0/0 or ∞/∞

lim_x→0 (x + 1)/(x + 2) = 1/2 by substitution. Differentiating would give 1, which is wrong.

12. Piecewise functions: checking only one formula

At the switch point, compute the left limit with the left formula and the right limit with the right formula. The limit exists only if they agree.

Practice section

1. lim_x→3 (x² − 9)/(x − 3)
2. lim_x→0 sin(3x)/(2x)
3. lim_x→∞ (2x³ − x)/(7x³ + 4)
4. lim_x→4 (√x − 2)/(x − 4)

Take the 8-question quiz below, then go deeper on the limits topic page. Not sure limits are your weakest topic? The free diagnostic tells you in about 5 minutes.