L'Hopital's rule
Top and bottom of a quotient both shrink to zero, or both grow without bound, and no algebra cancels them. Can the rates at which they move decide the limit instead?
Where you are 11 in Applications of derivatives
When you reach for it
Substitution into \dfrac{\ln x}{x^3 - 1} at x = 1 gives \dfrac{0}{0}. That is not an answer: a quotient whose top and bottom both go to 0 can approach any number, or none, depending on how fast each part shrinks. Limits by algebra: factoring and rationalizing settles such a limit by canceling a common factor, and The squeeze theorem and the limit of sine x over x settles \dfrac{\sin x}{x} with a squeeze. Neither helps here: there is no factor to cancel between a logarithm and a polynomial.
The limits \dfrac{0}{0} and \dfrac{\infty}{\infty}, and the shapes 0 \cdot \infty, \infty - \infty, 1^{\infty}, 0^0 and \infty^0 that can be rewritten as one of them, are called indeterminate forms: knowing the limits of the pieces does not determine the limit of the whole. L’Hopital’s rule compares the rates at which top and bottom change, which is what the derivatives measure.
The procedure
Day 1: the rule for zero over zero
L’Hopital’s rule. Let f and g be differentiable on an open interval I containing a, except possibly at a itself, with g'(x) \ne 0 for every x in I other than a. Suppose
If \lim_{x \to a} \dfrac{f'(x)}{g'(x)} equals a number L, or \infty, or -\infty, then \lim_{x \to a} \dfrac{f(x)}{g(x)} equals the same thing. The same holds for one-sided limits x \to a^+ or x \to a^- (with I an interval on that side), and for x \to \infty or x \to -\infty.
The rule takes the derivative of the top and of the bottom separately. It is not the quotient rule. And it runs one way: from a limit of \dfrac{f'}{g'} to the limit of \dfrac{f}{g}, never back.
Why it works: the simple case, by tangent lines. Suppose f(a) = g(a) = 0, both are differentiable at a, and g'(a) \ne 0. Near a each function is close to its linearization: f(x) \approx f'(a)(x - a) and g(x) \approx g'(a)(x - a), since f(a) = g(a) = 0. So the quotient is close to \dfrac{f'(a)}{g'(a)}, the ratio of the slopes, as the figure shows for \ln x and x^3 - 1.
The exact version uses the difference quotients. For x \ne a, divide top and bottom by x - a:
as x \to a, by the definition of the derivative and the quotient law for limits, which applies because the bottom limit g'(a) is not 0. (That bottom being near g'(a) \ne 0 also shows g(x) \ne 0 for x near a, so the quotient is defined.)
The general case needs a stronger mean value theorem. If g'(a) = 0, or f and g are not defined at a, the simple argument fails. The tool is the following, which we prove from Rolle’s theorem.
Cauchy’s mean value theorem. Let f and g be continuous on [a, b] and differentiable on (a, b), with g'(x) \ne 0 on (a, b). Then g(b) \ne g(a), and there is a number c in (a, b) with
Proof. If g(b) = g(a), Rolle’s theorem would give a point where g' = 0, which is ruled out. Now let
It is continuous on [a, b] and differentiable on (a, b), h(a) = 0, and h(b) = 0 because the two products are equal. Rolle’s theorem gives c with h'(c) = 0, that is, f'(c)\big[g(b) - g(a)\big] = g'(c)\big[f(b) - f(a)\big]. Divide by g'(c)\big[g(b) - g(a)\big], which is not 0. \blacksquare
With g(x) = x it is the ordinary mean value theorem.
Proof of the rule, x \to a^+. Give f and g the value 0 at a (changing a value at a does not change a limit as x \to a). Now both are continuous on [a, x] for x in I to the right of a, and Cauchy’s theorem gives a c_x with a < c_x < x and
As x \to a^+, c_x is trapped between a and x, so it is at least as close to a as x is, and \dfrac{f'(c_x)}{g'(c_x)} approaches L (or \pm\infty) because \dfrac{f'}{g'} does. The left side is the same with [x, a], and the two together give the two-sided limit. For x \to \infty, put \displaystyle x = \frac{1}{t} and let t \to 0^+: the chain rule gives \dfrac{d}{dt}\big[f(\tfrac{1}{t})\big] = -\dfrac{1}{t^2} f'(\tfrac{1}{t}), the factors -\dfrac{1}{t^2} cancel in the quotient of derivatives, and the one-sided case applies. \blacksquare
The rule may be used again on \dfrac{f'}{g'} if that is still \dfrac{0}{0}; the chain of equalities is then justified from the last limit backward.
Day 2: infinity over infinity, and the other forms
The rule for infinity over infinity. Keep the same hypotheses on f, g and g', but suppose f(x) \to \pm\infty and g(x) \to \pm\infty. If \dfrac{f'(x)}{g'(x)} tends to L, \infty or -\infty, so does \dfrac{f(x)}{g(x)}, for x \to a, one-sided, or x \to \pm\infty. This version is stated without proof here; its proof also rests on Cauchy’s theorem, applied on an interval [x, b] with b held fixed while x moves, but needs a more careful estimate. Its first use is a second proof of the ladder in Comparing growth rates at infinity (Example 4).
Products, 0 \cdot \infty. If f \to 0 and g \to \pm\infty, write fg = \dfrac{f}{1/g} (form \displaystyle \frac{0}{0}) or fg = \dfrac{g}{1/f} (form \displaystyle \frac{\infty}{\infty}). Both are correct; choose the one whose derivatives get simpler. A logarithm usually stays on top, since its derivative is simpler than it is.
Differences, \infty - \infty. Combine into one fraction, or factor out the larger term, until a quotient appears.
Powers, 1^{\infty}, 0^0 and \infty^0. For y = f(x)^{g(x)} with f(x) > 0, take logarithms as in logarithmic differentiation: \ln y = g(x) \ln f(x). In each of the three cases this is a 0 \cdot \infty form. If \ln y \to M, then y = e^{\ln y} \to e^M, because e^u is continuous. (If \ln y \to \infty then y \to \infty, and if \ln y \to -\infty then y \to 0.)
A limit for e. For every real number r,
Proof. For r = 0 both sides are 1. Otherwise, for large x the base is positive and \displaystyle \ln y = x \ln\left(1 + \frac{r}{x}\right) = \dfrac{\ln(1 + r/x)}{1/x}, a \displaystyle \frac{0}{0} form as x \to \infty. The quotient of derivatives is
so \ln y \to r and y \to e^r. \blacksquare With r = 1 this is the compounding limit of Compound interest and compounding periods, promised in the logarithm lesson.
When the rule is the wrong tool. The rule can loop, returning the quotient it started from, or make the quotient worse at every step (Example 7). And a quotient of polynomials at infinity is quicker by dividing by the highest power, as in Limits at infinity and horizontal asymptotes. Before each use, confirm the form is still indeterminate, and ask whether algebra is faster.
Worked examples
Day 1: Iracema’s logarithm over a cubic, and a limit done twice
(a) Find \lim_{x \to 1} \dfrac{\ln x}{x^3 - 1}. (b) Limits by algebra: factoring and rationalizing found \lim_{x \to 2} \dfrac{x^3 - 8}{x^2 - 4} = 3 by factoring. Iracema redoes it with the rule. Compare.
Solution(a) At x = 1 the top is \ln 1 = 0 and the bottom is 0: the form is \displaystyle \frac{0}{0}. Both functions are differentiable near 1, and g'(x) = 3x^2 \ne 0 there. The quotient of derivatives is continuous at 1:
This is the simple case: f'(1) = 1 and g'(1) = 3 are the slopes of the tangent lines in the figure. Check: at x = 1.001 the quotient is about 0.3328.
(b) The form is \displaystyle \frac{0}{0} and g'(x) = 2x \ne 0 near 2, so the limit is \lim_{x \to 2} \dfrac{3x^2}{2x} = \dfrac{12}{4} = 3. The two methods agree, as they must. Factoring explains why the form was \displaystyle \frac{0}{0} (a common factor x - 2); the rule needs no factoring and so also works where there is no factor, as in (a).
Day 1: Jurema’s linearization error, using the rule twice
Find \lim_{x \to 0} \dfrac{e^x - 1 - x}{x^2} and say what it means.
SolutionAt x = 0 the top is 1 - 1 - 0 = 0 and the bottom is 0. The bottom’s derivative 2x is not 0 for x \ne 0. The quotient of derivatives is \dfrac{e^x - 1}{2x}, again \displaystyle \frac{0}{0} at 0, so its limit is not yet known. Apply the rule to it (g'(x) = 2, never 0):
Read backward: this limit exists, so \lim_{x \to 0} \dfrac{e^x - 1}{2x} = \dfrac{1}{2}, and therefore
Check with a calculator: at x = 0.1 the quotient is about 0.5171, and at x = 0.01 about 0.5017.
The meaning: 1 + x is the linearization of e^x at 0, so the top is the error E(x) of that estimate. The limit says \displaystyle E(x) \approx \frac{1}{2}x^2 for small x, which is the “error proportional to (x - a)^2” of Linear approximation and differentials, now with its constant.
Day 1: when the hypotheses fail
(a) Moacir writes \lim_{x \to 1} \dfrac{x^2 + 1}{x - 1} = \lim_{x \to 1} \dfrac{2x}{1} = 2. Find the error and the correct answer. (b) Show that \lim_{x \to \infty} \dfrac{x + \sin x}{x} exists, though \lim_{x \to \infty} \dfrac{f'(x)}{g'(x)} does not.
Solution(a) The top tends to 2, not 0: the form is \displaystyle \frac{2}{0}, which is not indeterminate, and the rule does not apply. For x slightly above 1 the quotient is 2 divided by a small positive number, so it tends to \infty; slightly below 1 it tends to -\infty. The two-sided limit does not exist. The rule’s answer, 2, is wrong.
(b) Both top and bottom go to \infty, so this uses the \displaystyle \frac{\infty}{\infty} version of the rule, stated on Day 2 below, with the same hypothesis that \dfrac{f'}{g'} has a limit. Here \dfrac{f'(x)}{g'(x)} = \dfrac{1 + \cos x}{1} = 1 + \cos x, which swings between 0 and 2 forever and has no limit. The rule says nothing. But algebra works:
since \left|\dfrac{\sin x}{x}\right| \le \dfrac{1}{x}, the squeeze of The squeeze theorem and the limit of sine x over x. A missing limit of \dfrac{f'}{g'} never shows that \dfrac{f}{g} has no limit.
Day 2: Moacir’s second proof of the growth ladder
Show that for p > 0 and a > 1, and every positive integer n,
In the first, top and bottom go to \infty and g'(x) = p x^{p - 1} > 0 for x > 0. The quotient of derivatives is
so the first limit is 0.
In the second, the derivative of a^x is a^x \ln a, with \ln a > 0. Each use of the rule lowers the power on top by one and puts one more factor \ln a on the bottom:
Every quotient before the last is \displaystyle \frac{\infty}{\infty}, so each use is allowed, and the last one tends to 0 since a^x \to \infty. Read backward, \dfrac{x^n}{a^x} \to 0. For a power x^p that is not an integer, stop once the exponent turns negative: the top then goes to 0. These are the rungs of the ladder \ln x \ll x^p \ll a^x that Comparing growth rates at infinity proved with substitutions.
Day 2: Iracema’s product and difference
Find (a) \lim_{x \to \infty} x \sin \dfrac{3}{x} and (b) \lim_{x \to 1} \left(\dfrac{1}{\ln x} - \dfrac{1}{x - 1}\right).
Solution(a) As x \to \infty, x \to \infty and \displaystyle \sin \frac{3}{x} \to 0: the form is \infty \cdot 0. Move x to the bottom as \displaystyle \frac{1}{x}:
This is a \displaystyle \frac{0}{0} form.
By the chain rule the quotient of derivatives is \dfrac{\cos(3/x)\cdot(-3/x^2)}{-1/x^2} = 3\cos\dfrac{3}{x} \to 3, so the limit is 3. The other arrangement, \dfrac{x}{\csc(3/x)}, gives the quotient of derivatives \dfrac{x^2}{3\csc(3/x)\cot(3/x)}, which is harder than the start.
(b) As x \to 1^+ both fractions go to \infty: the form is \infty - \infty. Combine over the common denominator:
This is a \displaystyle \frac{0}{0} form. The quotient of derivatives is \dfrac{1 - 1/x}{\ln x + (x - 1)/x}. Multiply its top and bottom by x (allowed, since x \ne 0) to get \dfrac{x - 1}{x \ln x + x - 1}, again \displaystyle \frac{0}{0}. The rule once more gives
Reading backward, the limit is \dfrac{1}{2} from both sides. Check: at x = 1.01 the difference is about 0.4992.
Day 2: Jurema’s two power forms
Find (a) \lim_{x \to 0} (\cos x)^{1/x^2} and (b) \lim_{x \to 0^+} (\tan x)^x.
Solution(a) The base tends to 1 and the exponent to \infty: the form 1^{\infty}. Near 0, \cos x > 0, so let y = (\cos x)^{1/x^2} and take logarithms:
This is a \displaystyle \frac{0}{0} form. The quotient of derivatives is \dfrac{-\tan x}{2x}, still \displaystyle \frac{0}{0}; once more gives \dfrac{-\sec^2 x}{2} \to -\dfrac{1}{2}. So \displaystyle \ln y \to -\frac{1}{2} and
Check: at x = 0.1 the power is about 0.6060.
(b) The base tends to 0 and the exponent to 0: the form 0^0. For small x > 0, \tan x > 0. Let y = (\tan x)^x; then
a \displaystyle \frac{-\infty}{\infty} form. The quotient of derivatives is
So \ln y \to 0 and (\tan x)^x \to e^0 = 1. The form 0^0 does not always give 1 (exercise 23); here it does.
Day 2: a rule that loops and a rule that makes things worse
Find (a) \lim_{x \to \infty} \dfrac{e^x}{e^x + e^{-x}} and (b) \lim_{x \to 0^+} \dfrac{e^{-1/x}}{x}.
Solution(a) The form is \displaystyle \frac{\infty}{\infty}. The rule gives \dfrac{e^x}{e^x - e^{-x}}, and again gives \dfrac{e^x}{e^x + e^{-x}}, the starting quotient. It loops. Divide top and bottom by e^x instead:
(b) As x \to 0^+, \displaystyle -\frac{1}{x} \to -\infty, so the top goes to 0: the form is \displaystyle \frac{0}{0}. The rule gives \dfrac{e^{-1/x}\cdot(1/x^2)}{1} = \dfrac{e^{-1/x}}{x^2}, a higher power of x on the bottom, and every further use raises it again. Substitute \displaystyle t = \frac{1}{x} instead, so t \to \infty:
by Example 4 with n = 1 (or one use of the rule, \displaystyle \frac{1}{e^t} \to 0). The limit is 0.
Boundaries
- The form must be \displaystyle \frac{0}{0} or \displaystyle \frac{\infty}{\infty}. On any other form, such as \displaystyle \frac{2}{0} or \displaystyle \frac{1}{3}, the rule can give a wrong number (Example 3(a)). The other indeterminate forms must be rewritten as a quotient first.
- f and g must be differentiable near a (except possibly at a), and g'(x) \ne 0 for x near a, x \ne a.
- The limit of \dfrac{f'}{g'} must exist (or be \pm\infty). If it does not, the rule says nothing either way: the original limit may still exist (Example 3(b)).
- The rule is one-directional: it never shows a limit fails to exist.
- In a power form, the base must be positive for the logarithm to be defined.
Where this goes next
The rule is a statement about derivatives standing in for functions near a point. Taken further, a function is approximated not by its tangent line but by a polynomial matching several derivatives, which is where repeated use of the rule leads: Taylor and Maclaurin series and Using series to evaluate limits and integrals evaluate the same limits by comparing series term by term. The limits \displaystyle \left(1 + \frac{r}{x}\right)^x \to e^r and \ln x \ll x^p \ll a^x return wherever growth is compared or interest compounds without pause.
Exercises
Day 1: zero over zero
- \lim_{x \to 0} \dfrac{e^{4x} - 1}{\sin 2x}
- \lim_{x \to 1} \dfrac{x^5 - 1}{\ln x}
- \lim_{x \to 0} \dfrac{\tan^{-1}(3x)}{x}
- \lim_{x \to 0} \dfrac{x - \sin x}{x^3}
- \lim_{x \to 0} \dfrac{\ln(1 + x) - x}{x^2}
- \lim_{x \to 1} \dfrac{x^{1/3} - 1}{x^{1/2} - 1}
- \lim_{x \to \infty} \dfrac{\ln(1 + 3/x)}{\sin(1/x)}
- Find \lim_{x \to 0} \dfrac{\cos x}{x^2 + 1}, first deciding whether the rule applies. What would the rule have given?
- Find the numbers a and b for which \lim_{x \to 0} \dfrac{\sin 3x + ax + bx^3}{x^3} = 0.
Day 2: infinity over infinity, products and differences
- \lim_{x \to \infty} \dfrac{\ln(x^2 + 1)}{\ln x}
- \lim_{x \to \infty} \dfrac{e^{2x}}{x^2 + x}
- \lim_{x \to \infty} x\big(e^{1/x} - 1\big)
- \lim_{x \to \pi/2} \left(x - \dfrac{\pi}{2}\right)\tan x
- \lim_{x \to 0} \left(\dfrac{1}{\sin x} - \dfrac{1}{x}\right)
- \lim_{x \to \infty} (x - \ln x)
Day 2: power forms
- \lim_{x \to \infty} \left(1 - \dfrac{3}{x}\right)^{2x}
- \lim_{x \to 0} (1 + \sin 4x)^{\cot x}
- \lim_{x \to \infty} (e^x + x)^{1/x}
- \lim_{x \to 0^+} x^{\sin x}
Day 2: choosing a method
- Show that the rule loops on \displaystyle \lim_{x \to \infty} \dfrac{x}{\sqrt{x^2 + 9}}, then find the limit another way.
- Explain why the rule does not help with \lim_{x \to \infty} \dfrac{5^x}{4^x + 3^x}, and find the limit.
Reasoning
- Let f(x) = x^2 \sin \dfrac{1}{x} and g(x) = \sin x. Show that \lim_{x \to 0} \dfrac{f(x)}{g(x)} = 0 but \lim_{x \to 0} \dfrac{f'(x)}{g'(x)} does not exist. Does this contradict the rule?
- Show that for any real number k, \lim_{x \to 0^+} x^{k/\ln x} = e^k, and explain why this makes 0^0 an indeterminate form.
- Suppose f'' is continuous on an open interval containing a. Prove that \lim_{h \to 0} \dfrac{f(a + h) - 2f(a) + f(a - h)}{h^2} = f''(a).
Answers
Day 1: zero over zero
- 2. Form \displaystyle \frac{0}{0}; \dfrac{4e^{4x}}{2\cos 2x} \to \dfrac{4}{2} = 2.
- 5. Form \displaystyle \frac{0}{0}; \dfrac{5x^4}{1/x} = 5x^5 \to 5.
- 3. Form \displaystyle \frac{0}{0}; \dfrac{3/(1 + 9x^2)}{1} \to 3 (Derivatives of inverse trigonometric functions).
- \displaystyle \frac{1}{6}. Three uses: \dfrac{1 - \cos x}{3x^2}, then \dfrac{\sin x}{6x}, then \dfrac{\cos x}{6} \to \dfrac{1}{6}; the first two are still \displaystyle \frac{0}{0}.
- \displaystyle -\frac{1}{2}. \dfrac{1/(1 + x) - 1}{2x} is \displaystyle \frac{0}{0}; again, \dfrac{-1/(1 + x)^2}{2} \to -\dfrac{1}{2}.
- \displaystyle \frac{2}{3}. \displaystyle \dfrac{\frac{1}{3}x^{-2/3}}{\frac{1}{2}x^{-1/2}} \to \dfrac{1/3}{1/2} = \dfrac{2}{3}.
- 3. Form \displaystyle \frac{0}{0} at infinity; \dfrac{(-3/x^2)/(1 + 3/x)}{-\cos(1/x)/x^2} = \dfrac{3}{(1 + 3/x)\cos(1/x)} \to 3.
- 1. The form is \displaystyle \frac{1}{1}, not indeterminate; the function is continuous at 0, so substitute. The rule would have given \lim_{x \to 0} \dfrac{-\sin x}{2x} = -\dfrac{1}{2}, which is wrong.
- a = -3, \displaystyle b = \frac{9}{2}. The top tends to 0, so the rule gives \dfrac{3\cos 3x + a + 3bx^2}{3x^2}; for a finite limit its top must tend to 0, so 3 + a = 0. With a = -3, the rule twice more gives \dfrac{-27\cos 3x + 6b}{6} \to \dfrac{-27 + 6b}{6}, which is 0 when b = \dfrac{9}{2}.
Day 2: infinity over infinity, products and differences
- 2. Form \displaystyle \frac{\infty}{\infty}; \dfrac{2x/(x^2 + 1)}{1/x} = \dfrac{2x^2}{x^2 + 1} \to 2.
- \infty (no limit as a number). \dfrac{2e^{2x}}{2x + 1}, then \dfrac{4e^{2x}}{2} \to \infty.
- 1. \dfrac{e^{1/x} - 1}{1/x}, form \displaystyle \frac{0}{0}; \dfrac{e^{1/x}(-1/x^2)}{-1/x^2} = e^{1/x} \to 1.
- -1. \dfrac{x - \pi/2}{\cot x}, form \displaystyle \frac{0}{0}; \dfrac{1}{-\csc^2 x} \to -1.
- 0. \dfrac{x - \sin x}{x \sin x}; the rule gives \dfrac{1 - \cos x}{\sin x + x\cos x}, still \displaystyle \frac{0}{0}, then \dfrac{\sin x}{2\cos x - x\sin x} \to \dfrac{0}{2} = 0.
- \infty. Factor: x - \ln x = x\left(1 - \dfrac{\ln x}{x}\right), and \dfrac{\ln x}{x} \to 0 (rule: \dfrac{1/x}{1}), so the product goes to \infty.
Day 2: power forms
- e^{-6}. \displaystyle \left(1 - \frac{3}{x}\right)^{2x} = \Big[\left(1 + \frac{-3}{x}\right)^x\Big]^2 \to (e^{-3})^2.
- e^4. \ln y = \dfrac{\ln(1 + \sin 4x)}{\tan x}, form \displaystyle \frac{0}{0}; \dfrac{4\cos 4x/(1 + \sin 4x)}{\sec^2 x} \to 4.
- e. \ln y = \dfrac{\ln(e^x + x)}{x}; the rule gives \dfrac{e^x + 1}{e^x + x} = \dfrac{1 + e^{-x}}{1 + xe^{-x}} \to 1, since xe^{-x} \to 0.
- 1. \ln y = \sin x \ln x = \dfrac{\ln x}{\csc x}; \dfrac{1/x}{-\csc x \cot x} = -\dfrac{\sin x}{x}\tan x \to -1 \cdot 0 = 0, so y \to e^0.
Day 2: choosing a method
- 1. The rule gives \displaystyle \dfrac{1}{x/\sqrt{x^2 + 9}} = \dfrac{\sqrt{x^2 + 9}}{x}, and again gives \displaystyle \dfrac{x}{\sqrt{x^2 + 9}}. For x > 0, \displaystyle \dfrac{x}{\sqrt{x^2 + 9}} = \dfrac{1}{\sqrt{1 + 9/x^2}} \to 1.
- \infty. Each use multiplies the terms by \ln 5, \ln 4, \ln 3 and leaves the same exponentials, so nothing simplifies. Divide by 4^x: \dfrac{(5/4)^x}{1 + (3/4)^x} \to \dfrac{\infty}{1}.
Reasoning
- \dfrac{f(x)}{g(x)} = \dfrac{x}{\sin x}\cdot x\sin\dfrac{1}{x} \to 1 \cdot 0 = 0, since \displaystyle \left|x \sin \frac{1}{x}\right| \le |x|. But \dfrac{f'(x)}{g'(x)} = \dfrac{2x\sin(1/x) - \cos(1/x)}{\cos x}, and \displaystyle \cos \frac{1}{x} swings between -1 and 1, so it has no limit. No contradiction: the rule assumes \dfrac{f'}{g'} has a limit and says nothing when it does not.
- \ln y = \dfrac{k}{\ln x}\cdot \ln x = k for 0 < x < 1, so y = e^k for all such x. The base tends to 0 and the exponent \dfrac{k}{\ln x} tends to 0, so every one is a 0^0 form, yet the limits e^k take every positive value. The limits of the base and exponent do not decide the limit.
- Top and bottom tend to 0 (f is continuous) and the bottom’s derivative 2h is not 0 for h \ne 0. By the chain rule the quotient of derivatives in h is \dfrac{f'(a + h) - f'(a - h)}{2h}, again \displaystyle \frac{0}{0}; once more gives \dfrac{f''(a + h) + f''(a - h)}{2} \to \dfrac{2f''(a)}{2} by continuity of f''. Reading backward, the limit is f''(a).