L'Hopital's Rule

The last lesson left you with a wish. Instead of hunting for the right factor, the right conjugate, or the right series every single time, could you not just differentiate the top and bottom and try the limit again? That wish has a name, and it works.

A limit comes out 0/0 and you are stuck. Or ∞/∞. Both the numerator and the denominator are racing toward zero (or toward infinity), and plugging in tells you nothing about which one wins. L’Hopital’s rule is the escape hatch: in that exact situation, replace the top and bottom by their derivatives and take the limit again. Whichever side is changing faster wins the race, and the slopes — the derivatives — are what report the winner.

That “who is winning the race” question is the same one you ask when comparing growth rates — does n log n or n² dominate, does one loss term decay faster than another — so the instinct carries well past limit problems. But before any of it, one habit: check the form first. L’Hopital is licensed only for 0/0 and ∞/∞. Point it at 2/0 or 5/3 and it hands you 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 anywhere here. Second, if the new quotient f'/g' is still 0/0 or ∞/∞, you simply apply the rule again, and keep going until the form resolves.

That “differentiate the top and bottom” step is really a comparison of 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 (whose slope is 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, because L’Hopital is invalid otherwise and the wrong answer is built from forgetting that.

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       1

The exponential staple is identical in spirit — (e⁰ − 1)/0 = 0/0:

     eˣ − 1          eˣ        e⁰
lim  ──────  =  lim  ──  =     ──  =  1
x→0     x        x→0  1         1

Now the stubborn case, the one 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          2

So 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 only then read off the answer — the discipline is to re-check the form after every application, not to stop at the first.

A question to carry forward

Notice what these last three lessons had in common: every interesting limit was one where the function did not simply equal its plugged-in value — a hole, a jump, a 0/0 that needed work. But the easy, well-behaved case is worth a name too. When a curve flows smoothly through a point — when the limit as you approach equals the actual value f(a), with no hole and no jump — something special is true, and a great deal of later mathematics quietly depends on it. Here is the thread onward: what exactly does it mean for a function to be continuous at a point, and why is “the limit equals the value” the precise condition that guarantees no breaks?

In one breath

• L’Hopital: if lim f/g is 0/0 or ∞/∞, then lim f/g = lim f'/g' — differentiate top and bottom separately, then re-evaluate.
• Check the form first. Only 0/0 and ∞/∞ qualify; on 5/0 or 5/3 the rule gives a wrong answer.
• Repeat while the form stays indeterminate: (1 − cos x)/x² → sin x/2x → cos x/2 = 1/2 needs two applications.
• It is not the quotient rule — there is no g². Slope-against-slope is the whole idea.
• Staples to know cold: sin x/x → 1, (eˣ − 1)/x → 1, tan x/x → 1, (1 − cos x)/x² → 1/2.

Practice