L'Hopital's Rule
A shortcut for limits stuck at 0/0 or ∞/∞: differentiate top and bottom separately and try again. A reliable source of GATE DA limit marks.
What you'll learn
- L'Hopital applies only to the indeterminate forms 0/0 and ∞/∞ — check the form first
- The rule differentiates numerator and denominator separately — it is NOT the quotient rule
- You may apply it more than once if the form stays indeterminate
- Working the staple limits sin(x)/x, (eˣ−1)/x, and (1−cos x)/x² to a number
Before you start
A limit comes out 0/0 and you’re stuck. Or ∞/∞. Both sides are racing, and
plugging in tells you nothing about who wins. L’Hopital’s rule is a
gloriously simple escape hatch: in that exact situation, swap the top and bottom
for their derivatives and try the limit again. Whichever is changing faster
wins, and the slopes report it.
That “who’s winning the race” question is the same one you ask when comparing
growth rates — does n log n or n² dominate, does a loss term decay faster than
another — so the instinct transfers well beyond limit problems.
One habit before anything else — check the form first. L’Hopital is licensed
only for 0/0 and ∞/∞. Apply it to 2/0 or 5/3 and you’ll get a confidently
wrong answer.
The rule
If lim f(x)/g(x) is of the form 0/0 or ∞/∞ (and g' is non-zero near the
point), then:
f(x) f'(x)
lim ──── = lim ─────
g(x) g'(x)Two cautions live inside that picture. First, you differentiate f and g
independently — there is no quotient-rule denominator g² here. Second, if the
new quotient f'/g' is still 0/0 or ∞/∞, you simply apply the rule again.
The “differentiate the top and bottom” step is really comparing slopes near the
trouble point. Pick sin x in the widget below, drag the point to x = 0, and
read off f'(0) = 1. That is exactly the f'/g' value L’Hopital reads when the
denominator is x (slope 1): 1/1 = 1, the famous lim sin x / x.
How GATE asks this
Almost always a NAT: you are handed a single-variable limit that evaluates to
0/0 (occasionally ∞/∞) and asked for its value, usually a clean number. The
drill is fixed — confirm the form, differentiate top and bottom, re-evaluate, and
repeat once or twice if needed. Occasionally an MCQ tests whether you even
recognise a form as indeterminate, since L’Hopital is invalid otherwise.
Worked example — the staple limits
Start with the most famous limit in calculus. Substituting x = 0 gives 0/0, so
L’Hopital is licensed:
sin x cos x cos 0 1
lim ───── = lim ───── = ───── = ─ = 1
x→0 x x→0 1 1 1The exponential staple is identical in spirit — (e⁰ − 1)/0 = 0/0:
eˣ − 1 eˣ e⁰
lim ────── = lim ── = ── = 1
x→0 x x→0 1 1Now a case that needs the rule twice. At x = 0, (1 − cos x)/x² is 0/0:
1 − cos x sin x ← still 0/0 at x = 0, apply again
lim ───────── = lim ─────
x→0 x² x→0 2x
cos x cos 0 1
= lim ───── = ───── = ─
x→0 2 2 2So lim_{x→0} (1 − cos x)/x² = 1/2. After the first application the form was still
0/0, so we differentiated once more and then read off the answer.