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Limits and L'Hospital's rule questions
A limit that gives 0/0 or ∞/∞ on direct substitution is not thereby undefined; substitution has simply failed to evaluate it. Two methods do evaluate it: replace each function by its series expansion, or differentiate numerator and denominator separately, which is l'Hôpital's rule.
7 original questions · 27 marks · the limits and l'hospital's rule notes · Further Pure 1
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State L'Hospital's rule, including the condition under which it may be applied.
Worked 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. The quotient rule plays no part.Evaluate the limit as x → 0 of (1 − cos x)/x².
Worked 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.Evaluate the limit as x → 0 of (ex − 1 − x)/x², by series and by L'Hospital's rule.
Worked answer
Series: ex = 1 + x + x²/2 + …, so the numerator is x²/2 + x³/6 + … and the quotient tends to 1/2. L'Hospital: (ex − 1)/(2x), still 0/0, then ex/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. The series route needed no differentiation at all.Given tan x = x + x³/3 + 2x⁵/15 + …, evaluate the limit as x → 0 of (tan x − x)/x³.
Worked 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'Hospital needs three rounds here, each one differentiating sec²x again. Once a series is handed to you in the question, use it.Find the limit as x → ∞ of (1 + 3/x)x.
Worked answer
Take logarithms: ln y = x ln(1 + 3/x) = ln(1 + 3/x)/(1/x), which is 0/0 as x → ∞. Applying L'Hospital, 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. Power forms must be logged before the rule can see a quotient.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.
Worked answer
L'Hospital 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.Show that x ln x → 0 as x → 0+, and hence find the limit of xx as x → 0+.
Worked answer
As it stands x ln x is of the form 0 × (−∞), which L'Hospital 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 = xx. 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 xx, A1 for ln y → 0, A1 for the second limit. Two moves make this work. A product is turned into a quotient so the rule applies, and the power is logged before any limit is taken. Quoting 00 = 1 as a reason begs the question, because 00 is itself indeterminate.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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