Questions on limits and L'Hôpital's rule

Question 1

State L'Hôpital's rule, including the condition under which it may be applied.

Accepted Answer

If f and g both tend to 0, or both tend to infinity, as x approaches a, then the limit of f/g equals the limit of f'/g', provided that second limit exists. Top and bottom are differentiated separately. B1 for the rule, B1 for the 0/0 or ∞/∞ condition.

Question 2

Evaluate the limit as x → 0 of (1 − cos x)/x².

Accepted Answer

Both parts vanish at 0, so apply the rule, giving sin x/(2x), still 0/0. A second application gives cos x/2 → 1/2. M1 for the first application, M1 for the second, A1 for the limit. Alternatively cos x = 1 − x²/2 + …, so the numerator is x²/2 + … and the quotient tends to 1/2 in one line.

Question 3

Quote the two standard limits, for k > 0, of x^ke^−x as x → ∞ and of x^k ln x as x → 0⁺, and give a one-sentence reason for each.

Accepted Answer

x^ke^−x → 0 as x → ∞: the exponential grows faster than any fixed power, since each unit step in x multiplies e^x by e but x^k by a factor tending to 1. x^k ln x → 0 as x → 0⁺: ln x tends to −∞ more slowly than any positive power of x tends to 0, so the power wins. B1 B1, one for each limit with its reason. Both may be quoted as results unless a question says 'show that', and both hold for every positive k, however small.

Question 4

Explain why a limit of the form 0/0 carries no verdict, giving three quotients that all take the form 0/0 as x → 0⁺ and have three different limits.

Accepted Answer

The common limit 0 says nothing about the rates at which numerator and denominator vanish, and it is the ratio of those rates that fixes the answer. sin x/x → 1, x²/x = x → 0, and x/x² = 1/x → ∞ as x → 0⁺, all of them 0/0 at x = 0. B1 for the reason in terms of rates, B1 for three examples with different limits. The third is why the question keeps to x → 0⁺: from the left 1/x → −∞, so the two-sided limit of 1/x does not exist. The same holds for ∞/∞.

Question 5

Evaluate the limit as x → 0 of (e^x − 1 − x)/x², by series and by L'Hôpital's rule.

Accepted Answer

Series: e^x = 1 + x + x²/2 + …, so the numerator is x²/2 + x³/6 + … and the quotient tends to 1/2. L'Hôpital: (e^x − 1)/(2x), still 0/0, then e^x/2 → 1/2. Both routes agree. M1 for the exponential series, A1 for the numerator, A1 for the limit, M1 for two applications of the rule, A1 for confirming it.

Question 6

Given tan x = x + x³/3 + 2x⁵/15 + …, evaluate the limit as x → 0 of (tan x − x)/x³.

Accepted Answer

The numerator is x³/3 + 2x⁵/15 + …, so dividing by x³ gives 1/3 + 2x²/15 + … → 1/3. M1 for using the given series, A1 for the quotient, A1 for the limit. L'Hôpital needs three rounds here, each one differentiating sec²x again.

Question 7

Find the limit as x → ∞ of (1 + 3/x)^x.

Accepted Answer

Take logarithms: ln y = x ln(1 + 3/x) = ln(1 + 3/x)/(1/x), which is 0/0 as x → ∞. Applying L'Hôpital, or expanding ln(1 + u) ≈ u − u²/2, gives ln y → 3. So the limit is e³ ≈ 20.09. M1 for taking logarithms, M1 for writing it as a quotient, M1 for applying the rule, A1 for ln y → 3, A1 for the limit.

Question 8

A student evaluates the limit as x → 0 of (x + sin x)/(x + tan x) by differentiating top and bottom with the quotient rule, and gets an answer of 0. Diagnose the error and give the correct limit.

Accepted Answer

L'Hôpital differentiates numerator and denominator separately, giving (1 + cos x)/(1 + sec²x) → 2/2 = 1. The quotient rule computes the derivative of the whole fraction, which answers a different question entirely. The correct limit is 1, and the series route agrees, since both parts are 2x + … near zero. B1 for naming the misuse of the quotient rule, M1 for differentiating numerator and denominator separately, A1 for the limit.

Question 9

Use L'Hôpital's rule to evaluate the limit as x → 0 of (sin x − x cos x)/x³.

Accepted Answer

At x = 0 both numerator and denominator are 0. Differentiating separately: the numerator gives cos x − cos x + x sin x = x sin x and the denominator 3x², so the quotient becomes x sin x/(3x²) = sin x/(3x). M1 for the first application, A1 for sin x/(3x). This is still 0/0; a second application gives cos x/3, or sin x/x → 1 can be quoted. M1. The limit is 1/3. A1. Cancelling the x before differentiating again keeps the second round to one line; differentiating x sin x/(3x²) as it stands gives (sin x + x cos x)/(6x), which is correct but still 0/0 and needs a third round. Series confirm it: x − x³/6 − x(1 − x²/2) = x³/3 + …

Question 10

Find the limit as x → 0 of 1/x − 1/sin x.

Accepted Answer

As it stands the form is ∞ − ∞, which the rule cannot handle. Put the difference over a common denominator: (sin x − x)/(x sin x), now 0/0. M1 for combining into one quotient, A1 for (sin x − x)/(x sin x). By series the numerator is −x³/6 + … and the denominator x² + …, so the quotient is −x/6 + … → 0; by the rule, (cos x − 1)/(sin x + x cos x) is still 0/0 and a second round gives −sin x/(2 cos x − x sin x) → 0/2 = 0. M1 for either route. The limit is 0. A1. The two reciprocals each grow without bound, so the answer 0 is not obvious in advance: they grow at the same rate to first order and differ only at order x.

Question 11

Find the limit as x → ∞ of ln x/√x, stating the indeterminate form first.

Accepted Answer

Both parts tend to infinity, so the form is ∞/∞ and the rule applies. B1 for naming the form. Differentiating separately gives (1/x)/(1/(2√x)) = 2√x/x = 2/√x. M1. That tends to 0, so the limit is 0. A1. This is the logarithm losing to a power at infinity, the mirror image of the standard limit x^k ln x → 0 at the origin, and putting x = 1/u turns one into the other.

Question 12

Show that x ln x → 0 as x → 0⁺, and hence find the limit of x^x as x → 0⁺.

Accepted Answer

As it stands x ln x is of the form 0 × (−∞), which L'Hôpital cannot touch, so rewrite it as the quotient x ln x = ln x/(1/x), now of the form −∞/∞. Differentiating top and bottom gives (1/x)/(−1/x²) = −x, which tends to 0. So x ln x → 0. For the second part let y = x^x. Then ln y = x ln x → 0, and since the exponential is continuous, y → e⁰ = 1. M1 for rewriting the product as a quotient, M1 for applying the rule, A1 for the first limit, M1 for taking logarithms of x^x, A1 for ln y → 0, A1 for the second limit. Quoting 0⁰ = 1 as a reason begs the question, because 0⁰ is itself indeterminate.

Question 13

Show that x³e^−x → 0 as x → ∞ by repeated use of L'Hôpital's rule, and explain why the same argument works for x^ke^−x with any positive integer k. By writing x = e^−u, deduce that x^k ln x → 0 as x → 0⁺ for every k > 0.

Accepted Answer

Write x³e^−x = x³/e^x, which is ∞/∞. Three rounds of the rule give 3x²/e^x, then 6x/e^x, then 6/e^x, and the last tends to 0. M1 for the quotient form and the rule, A1 for 6/e^x → 0. For a positive integer k, each round lowers the power by one and leaves e^x alone, so after k rounds the numerator is the constant k! while the denominator still grows without bound. B1 for that argument.
Now put x = e^−u, so that ln x = −u and u → ∞ as x → 0⁺. Then x^k ln x = −u e^−ku. M1 for the substitution with the new limit. With v = ku this is −(1/k)ve^−v, and v → ∞ as u does, so by the first result with k = 1 it tends to 0. A1 for the rearrangement, A1 for the conclusion x^k ln x → 0.

Question 14

Explain why L'Hôpital's rule cannot be applied to the limit as x → 0 of sin x/(x + 1), and state that limit. Then show that (x + sin x)/x → 1 as x → ∞, and explain why applying the rule to this quotient produces no conclusion, even though the form is ∞/∞.

Accepted Answer

At x = 0 the numerator is 0 but the denominator is 1, so the form is 0/1, not indeterminate, and the limit is simply 0. Applying the rule anyway would return cos x/1 → 1, which is wrong. B1 for the form and the limit.
For the second quotient, divide through: (x + sin x)/x = 1 + (sin x)/x, and since |sin x| is at most 1, (sin x)/x → 0 as x → ∞, so the limit is 1. M1 for dividing by x and bounding sin x, A1 for the limit.
The form is ∞/∞, so the rule may be tried, but differentiating separately gives (1 + cos x)/1, which oscillates between 0 and 2 for ever and has no limit. B1 for showing that f'/g' has no limit. The rule says that if lim f'/g' exists then lim f/g equals it; when lim f'/g' does not exist the rule is silent, and the original limit has to be found another way. B1 for the explanation. The rule never proves a limit fails to exist.

Question 15

Find the limit as x → 0 of (cos x)^1/x².

Accepted Answer

As x → 0 the base tends to 1 and the exponent to infinity, so this is the power form 1^∞, and logarithms come first. Let y = (cos x)^1/x²; then ln y = ln(cos x)/x², which is 0/0. M1 for taking logarithms, A1 for the quotient and its form.
Applying the rule: (−sin x/cos x)/(2x) = −tan x/(2x), still 0/0. M1. A second application gives −sec² x/2 → −1/2, or quote tan x/x → 1. M1. So ln y → −1/2. A1.
Since the exponential is continuous, y → e^−1/2, about 0.607. A1. By series, ln(cos x) = ln(1 − x²/2 + …) = −x²/2 + …, and dividing by x² gives −1/2 at once. Stopping at ln y → −1/2 answers a different question; the limit asked for is of y, and the final step of exponentiating has to be written.