x3 + x " 2 . x!1 x 2 " 3x + 2
This an attempt by a student to answer the question above. How many marks (out of 10) would you give this answer? What needs to change in order to gain full marks?
x3 + x " 2 x!1 x 2 " 3x + 2
3x 2 +1 x!1 2x " 3
= lim = lim x!1
= 3 by l'Hopital's rule.
The following extracts from the M203 Handbook may be helpful: a) Let f be defined on an open interval I, with a ! I . Then f is continuous at a if and only if lim f x = f a x!a
b) Polynomial functions are continuous and differentiable on R c) Composition Rules: If lim f (x) = l and lim g(x) = m , then x!a
Sum Rule: Multiple Rule:
( ( ) ( ))
lim f x + g x = l + m x!a
lim ! f (x) = ! l, for ! ! R x!a
Product Rule: lim f (x)g(x) = lm x!a
Quotient Rule: lim f (x) / g(x) = l / m, provided m " 0 x!a
d) L'H么pital's Rule: Let f and g be differentiable on an open interval I containing the point c, at which f (c) = g(c) = 0 . Then lim x!c
f (x) f '(x) exists and equals lim provided that this last limit exists. x!c g(x) g '(x)
It may be necessary to apply the Rule more than once in order to find a particular limit.
Marking time – notes for tutors This was a 45 minute session, led by a tutor in each breakout room. We didn't hand out solutions to the students – we assumed they'd make their own notes. This is the strategy the tutors followed in running the session.
Students are asked to mark this argument, and say what it would need to get full marks. x3 + x " 2 3x 2 +1 6x = lim = lim 2 x!1 x " 3x + 2 x!1 2x " 3 x!1 2
= 3 by L'Hôpital's Rule
This is what I had in mind for this question: What's wrong with it?
They should point out: • Criteria for L'Hôpital's Rule not satisfied at the second stage (lose 1 mark) • The answer is wrong (lose 1 mark) • No justification given at any stage. (lose 8 marks) [More to do with marks than rigour !] • So 0/10 How to fix it. A
• try evaluating the function at x = 1: gives 0/0 so think immediately of L'Hôpital • check both criteria at first stage - OK • check the criteria at second stage - not OK • so look for something else - try f'(1)/g'(1) = 4/(−1) = −4 - looks suspiciously continuous • can we prove it is continuous? part a) in the notes doesn't seem to help - need to know the limit! • easy to show separate bits are continuous, by part b) • then we know the limits of the separate bits by part a) - does this help? • composition rules look inviting - which rule? what criteria must be satisfied? B Strategy:
1 2 3 4
Check criteria for L'Hôpital Apply L'Hôpital's Rule Check criteria for quotient rule Apply quotient rule
C Writing down the proof: Notes
Let f (x) = x + x ! 2, and g(x) = x ! 3x + 2
Helps to structure the proof if you apply labels
Then f and g (polynomial functions) are differentiable on any open interval containing 1 and f(1) = g(1) = 0.
4 marks for checking criteria
Therefore we may use L'Hôpital's Rule to show that x3 + x " 2 f (x) f '(x) = lim = lim x!1 x 2 " 3x + 2 x!1 g(x) x!1 g '(x)
f '(x) = 3x 2 +1, g '(x) = 2x ! 3 are continuous at 1, so
lim f '(x) = f '(1) = 4, and lim g '(x) = g '(1) = "1 (# 0) x!1
1 mark for correct application of L'Hôpital's Rule 2 marks for finding limits 1 mark for checking final criterion for quotient rule
By the quotient rule, we have lim x!1
f '(x) f '(1) 4 f '(x) lim = x!1 = = = "4 g '(x) lim g '(x) g '(1) "1
1 mark for correct use of quotient rule
x3 + x " 2 = "4 x!1 x 2 " 3x + 2
1 mark for correct answer
The brighter students may notice that there is a removable discontinuity at 1, and factor out x âˆ’ 1. This is fine. They would still need to use the quotient rule on the result,
x2 + x + 2 , or find a way to prove it is continuous x!2
before finding the limit. Good opportunity to make the point (again!) that there may be more than one way to skin cat. Doesn't matter if they don't find the quick way - slogging through the details in the first method is equally acceptable, and probably good for their souls.