Taylor polynomials, question 12: UBC MATH 100 practice

Question 12 of 18Standard3 marks · 5 minTheory UBC Math 100 §9.5.2CLP-1 §3.4Key results

Use Maclaurin polynomials to evaluate the limit.

\lim_{x \to 0} \frac{e^{2x} - 1 - 2x - 2x^{2}}{x - \sin x}

Your answer

A whole number.

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Theory UBC Math 100 §9.5.2 CLP-1 §3.4Key results

Worked solution

Strategy

Replace each function by its Maclaurin polynomial, keeping terms only up to the power that decides the limit. The bottom starts at x³, so the top is needed up to x³ as well.

Working

$x - \sin x \approx x - \left(x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!}\right) = \frac{x^{3}}{6} - \frac{x^{5}}{120}$
The x terms cancel, so the bottom behaves like x³/6 near 0, and the top is needed up to x³.
$e^{2x} \approx 1 + 2x + \frac{(2x)^{2}}{2!} + \frac{(2x)^{3}}{3!} = 1 + 2x + 2x^{2} + \frac{4}{3}x^{3}$
Put u = 2x into the polynomial of e^u; each power applies to the 2 as well.
$e^{2x} - 1 - 2x - 2x^{2} = \frac{4}{3}x^{3} + (\text{higher powers})$
Everything up to x² cancels, leaving 4x³/3 as the lowest power.
$\lim_{x \to 0} \frac{\tfrac{4}{3}x^{3} + \cdots}{\tfrac{1}{6}x^{3} - \cdots} = \lim_{x \to 0} \frac{\tfrac{4}{3} + \cdots}{\tfrac{1}{6} - \cdots} = \frac{4/3}{1/6} = 8$
Divide the top and the bottom by x³; every higher power then tends to 0.

Answer

8

Check

Three applications of l’Hôpital’s rule reach 8e^2x/cos x, which is 8 at x = 0; and at x = 0.01 the quotient is about 8.04.

Where this goes wrong

Stopping the expansions too early. With e^2x only up to x², the top looks like 0 and the limit like 0/0 again; the order needed is set by the lowest power left in the bottom, here x³.

The idea behind it

The Taylor polynomials lesson, a figure to move, in a few steps

Leans on

Limits of the form 0/0, from L’Hôpital’s rule

Try a similar question

Find the coefficient of

x^{3}

in the third-order Maclaurin polynomial of

f(x) = e^{2x}\sin x

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Taylor polynomials practice question 12