When a limit has a form like 0 over 0 that does not decide its value, how l’Hôpital’s rule settles it, which hypotheses it needs, and how to handle products, differences and powers.
As x→0 both sinx and x shrink to 0, so their quotient is a race between two vanishing quantities. Knowing that both go to 0 says nothing about who wins. The derivatives measure how fast each one is shrinking, and l’Hôpital’s rule turns that idea into a precise tool for limits that algebra alone cannot open.
Key terms
The forms 0 over 0 and infinity over infinity
Definition Let f and g be defined near a, except possibly at a, with g(x)=0 there. The limit
x→alimg(x)f(x)
is of indeterminate type 00 if f(x)→0 and g(x)→0 as x→a. It is of type ∞∞ if ∣f(x)∣→∞ and ∣g(x)∣→∞ as x→a. Indeterminate means that the type alone does not determine the limit, which may be any real number, ±∞, or may not exist.
In plain words Zero over zero is a question, not an answer. Different functions of this type give different limits.
Example All three of these limits are of type 00 at 0.
x→0limxx=1,x→0limxx2=0,x→0+limx2x=∞.
The rule and its hypotheses
Definition This is l’Hôpital’s rule. Let I be an open interval containing a. Suppose f and g are differentiable on I except possibly at a, and g′(x)=0 for every x in I with x=a. Suppose the limit of f(x)/g(x) as x→a is of type 00 or ∞∞. If
x→alimg′(x)f′(x)=L,
where L is a real number, ∞ or −∞, then also
x→alimg(x)f(x)=L.
In the ∞∞ case it is in fact enough that ∣g(x)∣→∞, whatever f does.
In plain words For a 00 or ∞∞ quotient, you may replace top and bottom by their derivatives (separately, not with the quotient rule), as long as the new limit exists or is infinite and the bottom derivative is not zero near a.
Example The limit of (ex−1)/x as x→0 is of type 00, both functions are differentiable everywhere, and g′(x)=1=0. So
x→0limxex−1=x→0lim1ex=1.
The type check matters. The limit
x→0limx+2x+1=21
is not indeterminate, and differentiating anyway would give the wrong value 1.
One-sided limits and limits at infinity
Definition The rule holds with the same conclusion for one-sided limits x→a+ and x→a−, using an interval (a,a+δ) or (a−δ,a) in place of I, and for x→∞ or x→−∞, using an interval (N,∞) or (−∞,N). The limit L may be infinite in every case.
In plain words The rule works from one side and far out on the number line too, and it can show that a limit is infinite.
Example As x→∞ the quotient (lnx)/x is of type ∞∞, and g′(x)=1/(2x)=0 for x>0. So
x→∞limxlnx=x→∞lim1/(2x)1/x=x→∞limx2=0.
From the right at 0 the quotient sinx/(1−cosx) is of type 00, and g′(x)=sinx=0 on (0,π). The quotient of derivatives satisfies
x→0+limsinxcosx=∞,
so the original limit is ∞ too.
Indeterminate products
Definition If limx→af(x)=0 and limx→a∣g(x)∣=∞, the limit of f(x)g(x) is of indeterminate type 0⋅∞. Rewrite it as a quotient,
f(x)g(x)=1/g(x)f(x)orf(x)g(x)=1/f(x)g(x),
which is of type 00 or ∞∞. The first form needs g(x)=0 near a, which holds because ∣g(x)∣→∞. The second needs f(x)=0 for x near a with x=a.
In plain words Move one factor into the denominator as its reciprocal. Choose the one whose derivative stays simple.
Example For the limit of xlnx as x→0+, write the product as (lnx)/(1/x), of type ∞∞. Then
x→0+lim1/xlnx=x→0+lim−1/x21/x=x→0+lim(−x)=0.
Indeterminate differences
Definition If limx→af(x)=∞ and limx→ag(x)=∞, the limit of f(x)−g(x) is of indeterminate type ∞−∞. Rewrite the difference as a single quotient, by a common denominator, by rationalizing, or by factoring, and then evaluate that quotient.
In plain words Two huge quantities can cancel in any proportion. Combine them into one fraction first.
Example Consider the limit
x→0+lim(x1−sinx1),
where both terms grow without bound. A common denominator gives
x1−sinx1=xsinxsinx−x,
which is of type 00. Its value is found in the section on repeated use below.
Indeterminate powers
Definition A limit of f(x)g(x) with f(x)>0 near a is indeterminate when it is of type 00 (f→0+, g→0), ∞0 (f→∞, g→0) or 1∞ (f→1, ∣g∣→∞). Write
f(x)g(x)=eg(x)lnf(x).
Because the exponential function is continuous, if g(x)lnf(x)→M for a real number M, then
x→alimf(x)g(x)=eM.
If instead g(x)lnf(x)→∞, the power tends to ∞, and if it tends to −∞, the power tends to 0.
In plain words Take the logarithm to bring the exponent down, find that limit, then exponentiate the answer.
Example The limit limx→0+xx is of type 00. Here g(x)lnf(x)=xlnx→0 by the product example above, so
x→0+limxx=e0=1.
Using the rule more than once
Definition If f′/g′ is again of type 00 or ∞∞ and f′ and g′ satisfy the hypotheses, the rule may be applied to f′/g′, and so on. The hypotheses must be checked again at each step, and you must stop as soon as the quotient is no longer indeterminate.
In plain words Keep going while the form stays 00 or ∞∞. Check the form every time.
The difference example also needs two steps. Here the interval (0,π/4) plays the role of (a,a+δ). On it the two denominators that the rule differentiates, xsinx and then sinx+xcosx, have derivatives sinx+xcosx>0 and 2cosx−xsinx>0, so the hypothesis on g′ holds at both steps, and
Definition If f′(x)/g′(x) has no limit as x→a, finite or infinite, then l’Hôpital’s rule says nothing about the limit of f(x)/g(x), which may still exist. The rule can also go in circles, with each step giving back an equivalent form. In both cases use another method, such as algebra, bounds or the squeeze theorem.
In plain words A failed rule does not mean the limit fails to exist. It just means you need a different approach.
Example Take the quotient (x+sinx)/x as x→∞. The quotient of derivatives is 1+cosx, which keeps swinging between 0 and 2 and has no limit. Yet for x>0
xx+sinx=1+xsinx→1,becausexsinx≤x1.
Similarly, for x2+1/x the rule gives x/x2+1, and applying it again gives back the original. Algebra settles it at once, since for x>0
xx2+1=1+x21→1as x→∞.
Check yourself
Find limx→0xsin3x.
Show the answer
It is of type 00 with g′(x)=1, so
x→0limxsin3x=x→0lim13cos3x=3.
Find limx→∞x2e−x.
Show the answer
Write it as x2/ex, of type ∞∞, where the denominator has derivative ex=0. Two applications give
x→∞limexx2=x→∞limex2x=x→∞limex2=0.
Find limx→0+(1+2x)1/x.
Show the answer
It is of type 1∞. The exponent of e is ln(1+2x)/x, of type 00, and
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