Try
Near zero, e^x−1 and x both approach zero. Their quotient is not found by dividing zero by zero. You already know its limit is 1 from the exponential derivative at zero. This provides a known case against which to check a new theorem-based method.
Predict
A learner differentiates numerator and denominator of (x²+1)/(x+1) and reports limit zero at x=0. Direct substitution gives 1. Predict which missing hypothesis makes the derivative-ratio substitution invalid. Differentiable formulas alone are not enough.
Experiment
Classify the original limit before performing any differentiation. The entries “0/0” and “infinity/infinity” describe indeterminate behavior, not numerical answers.
| Original expression and approach | Form | Suitable first action |
|---|
| (e^x−1)/x at zero | 0/0 | check L’Hopital hypotheses |
| (x²+1)/(x+1) at zero | 1/1 | substitute directly |
| x ln x at zero from right | zero times unbounded | rewrite as a quotient |
| (x+sin x)/x at positive infinity | infinity/infinity | direct bounds are simpler |
For the last row, divide through by x to get 1+sin x/x. Its limit is 1 by a shrinking absolute bound. Differentiating instead gives 1+cos x, which has no limit. A valid original limit can exist even when this method becomes inconclusive.
Understand
In a standard local form of L’Hopital’s rule, f and g are differentiable on a punctured interval approaching the target, g′ is nonzero there, and the quotient is of form 0/0 or infinity/infinity. If f′/g′ has a finite or appropriate infinite limit, the original quotient has that limit. One-sided and infinite-input versions use corresponding intervals. Check that the original quotient is defined on the approach as well. The theorem relates limits under hypotheses; it does not say f/g equals f′/g′ as functions.
For (e^x−1)/x near zero, the derivative ratio is e^x/1→1. For (1−cos x)/x², the first derivative ratio is sin x/(2x), still 0/0. A second application gives cos x/2→1/2. Before each application, check the new form and denominator derivative. The first denominator derivative 2x is nonzero on a punctured neighborhood; the second is 2. Radian trigonometric formulas are assumed.
Prove the rule: secant ratios and endpoint limits
The precise claim. Let I be an open interval and let c be one of its endpoints, possibly positive or negative infinity. Approach c from inside I. Suppose f and g are differentiable on I, g′(x)≠0 there, and f′(x)/g′(x) tends to L, where L is a finite real number, positive infinity or negative infinity. Either of the following assumptions suffices:
- f(x)→0 and g(x)→0 as x→c;
- |g(x)|→infinity as x→c (no separate assumption on the limit of f is needed).
Then f(x)/g(x) is defined sufficiently near c and tends to L. We may shrink I to a terminal subinterval throughout: for a finite endpoint this is a sufficiently short one-sided interval; at positive infinity it is a tail (R,infinity), and at negative infinity a tail (−infinity,R). A two-sided finite-endpoint conclusion requires the hypotheses on both sides and the same resulting limit.
Why a secant ratio equals a derivative ratio somewhere. Take distinct x,y in I. On the closed interval between them define
h(t)=(f(x)−f(y))(g(t)−g(y))−(g(x)−g(y))(f(t)−f(y)).
This function is continuous on the closed interval, differentiable inside, and zero at both endpoints. Rolle’s theorem gives an interior point ξ with h′(ξ)=0. Also g(x)−g(y)≠0: otherwise Rolle applied to g would give a zero of g′. Rearranging h′(ξ)=0, and using g′(ξ)≠0, gives
(f(x)−f(y))/(g(x)−g(y)) = f′(ξ)/g′(ξ).
This is Cauchy’s mean value theorem in the ratio form needed here. No continuity of the derivatives is assumed. Whenever the derivative ratio lies between two bounds throughout a terminal subinterval, every secant ratio with both endpoints in that subinterval lies between those same bounds. A one-sided bound transfers in exactly the same way.
Case 1: both functions tend to zero. First, g cannot vanish anywhere on I. If g(x)=0 at an interior x while its limit at endpoint c is also zero, choose an interior z between x and c. If g(z)=0, Rolle already contradicts g′≠0. Otherwise choose y closer to c than z with |g(y)|<|g(z)|. On the closed interval with endpoints x,y, g has an interior maximum or minimum, since g(z) lies strictly beyond both endpoint values. Fermat’s theorem again gives g′=0, a contradiction.
For finite L and ε>0, choose a terminal subinterval where L−ε<f′/g′<L+ε. Fix x there and let y approach c within it. Cauchy’s theorem bounds the secant ratio; its limit as y→c is f(x)/g(x), so
L−ε ≤ f(x)/g(x) ≤ L+ε.
Using ε/2 for any requested error proves the limit. This also works when c is infinite: the argument uses endpoint limits, not a value assigned at c. If L=positive infinity, start with any bound M, take f′/g′>M+1 near c, and pass y→c to obtain f/g≥M+1>M. For negative infinity, use f′/g′<M−1 for any prescribed upper threshold M. Thus both infinite limits follow without dividing or multiplying inequalities by an unknown-sign denominator.
Case 2: the denominator grows without bound. For finite L and ε>0 choose a terminal subinterval where |f′/g′−L|<ε, and fix one y inside it. For x closer to c than y, let R(x,y) be the secant ratio. Cauchy’s theorem gives |R(x,y)−L|<ε. For x sufficiently near c, g(x)≠0 and the exact identity is
f(x)/g(x) = R(x,y)(1−g(y)/g(x)) + f(y)/g(x).
Subtract L. The absolute error is at most ε|1−g(y)/g(x)|+|f(y)−L g(y)|/|g(x)|. The first factor tends to 1 and the second term to zero, because y is fixed before x approaches c. For a desired error δ>0 choose ε=δ/4, then choose x so the factor is below 2 and the second term below δ/2. This proves the finite limit. It also explains why a derivative-ratio statement about increments does not immediately prove a quotient limit: the fixed-endpoint terms must disappear.
If L=positive infinity, fix any desired lower threshold M. Choose K>2(|M|+1) and then a terminal subinterval where f′/g′>K. Fix y there. Eventually R(x,y)>K, 1−g(y)/g(x)>1/2, and |f(y)/g(x)|<1. The identity gives f(x)/g(x)>K/2−1>M. For L=negative infinity, apply this positive-infinity argument to −f and g, then negate the conclusion. This proves the unbounded-denominator versions, including the usual infinity/infinity applications.
Practice the proof mechanism. Explain why y must remain fixed in Case 2, but tends to c in Case 1. Then prove log(x)/x^r→0 as x→positive infinity for an arbitrary fixed r>0: the denominator grows without bound, its derivative r x^(r−1) is positive, and the derivative ratio is 1/(r x^r)→0. For a changed boundary, prove x^r log(x)→0 as x→0 from the right by using g(x)=x^(−r). Its derivative ratio is −x^r/r→0. Identify which part of the theorem justifies each step; a numerical graph cannot prove the assertion for every r>0.
Other indeterminate forms require transformation. For x ln x as x approaches zero from the right, write ln x/(1/x). The derivative ratio is (1/x)/(−1/x²)=−x→0. The right-hand approach is essential because ln x is real only for positive inputs. For (1+x)^(1/x), take logarithms with 1+x positive. The logarithm is ln(1+x)/x→1, so continuity of the exponential gives the original limit e.
If the derivative ratio has no limit, the rule supplies no conclusion. It is not a reversible test for whether the original limit exists. Nor is it valid for a determinate ratio such as 1/1, or without the differentiability and nonzero derivative conditions. Algebra, squeeze bounds and known limits often remain the clearer methods.
Apply
Write a short eligibility check before each practice application: approach and domain, original form, differentiability, denominator derivative, derivative-ratio limit. In the counterexample task, prove the original limit independently. For a variable power, compute the logarithmic limit first and explain the final exponentiation.
Continue
This completes a differential calculus unit covering techniques and applications, not full Calculus I or AP AB/BC. Integral calculus, further approximation and cumulative assessment remain separate work. Revisit unfamiliar practice without treating previously revealed solutions as fresh independent evidence.
