0/0 is a question, not an answer: Picture sliding x in towards a value and watching a fraction. If the top → 0 and the bottom → 0 at the same time, the fraction is a tug-of-war: the answer could be anything.
That shape 0/0 (or ∞/∞) is called an indeterminate form — it just means "substitution failed, work harder." It is NOT 0, and NOT undefined; the limit may well exist.
First, always try substitution: Before any clever trick, put the number in.
If you get a clean number, that's your limit — done.
Only if you hit 0/0 or ∞/∞ do you reach for L'Hopital's rule.
IB-style question — is it indeterminate?
A student wants the limit of (eˣ − 1)/x as x approaches 0.
Show that substituting x = 0 gives an indeterminate form.
Step by step
- Substitute x = 0 into the top: e⁰ − 1 = 1 − 1 = 0.
- Substitute x = 0 into the bottom.
- Top → 0 and bottom → 0, so the form is the indeterminate 0/0.
Final answer
Both top and bottom → 0, so it is the indeterminate form 0/0 — substitution alone cannot decide the limit.
Differentiate top and bottom on their own: If lim f(x)/g(x) gives 0/0 or ∞/∞, then
lim f(x)/g(x) = lim f′(x)/g′(x).
Why it works: near the trouble point both curves pass through the same height, so the ratio of the fractions is governed by the ratio of their slopes. The faster-falling one wins.
The #1 trap: this is NOT the quotient rule. Differentiate the top by itself and the bottom by itself, then divide.
IB-style question — apply the rule once
Find the limit of (eˣ − 1)/x as x approaches 0.
Step by step
- Substitution gives 0/0 (shown earlier), so L'Hopital applies.
- Differentiate the top on its own: d/dx(eˣ − 1) = eˣ.
- Differentiate the bottom on its own: d/dx(x) = 1.
- Take the new limit by substituting x = 0.
Final answer
The limit is 1.
IB-style question — a trig limit
Find the limit of (sin x)/x as x approaches 0.
Step by step
- Substitution: sin 0 / 0 = 0/0, so use L'Hopital.
- Top derivative: d/dx(sin x) = cos x. Bottom derivative: d/dx(x) = 1.
- Substitute x = 0.
Final answer
The limit is 1 — the famous result that sin x ≈ x for small x.