Composites Done, Now What?

Calculus Level 3

$\large\lim_{x\to 0}\dfrac{\tan x - x}{x^2\tan x}$

The limit above has a closed form. Find the value of this closed form.

Give your answer to 3 decimal places.

The answer is 0.333.

1 solution

Chew-Seong Cheong
May 28, 2016

Relevant wiki: Maclaurin Series

$\begin{aligned} \mathfrak L & = \lim_{x \to 0} \frac{\color{#3D99F6}{\tan x} - x}{x^2\color{#3D99F6}{\tan x}} \quad \quad \small \color{#3D99F6}{\text{Using Maclaurin series}} \\ & = \lim_{x \to 0} \frac{\color{#3D99F6}{x+\frac{1}{3}x^3 +\frac{2}{15}x^5 + \frac{17}{315}x^7+ \cdots} - x}{x^2 \left( \color{#3D99F6}{x+\frac{1}{3}x^3 +\frac{2}{15}x^5 + \frac{17}{315}x^7+ \cdots}\right)} \\ & = \lim_{x \to 0} \frac{\frac{1}{3}x^3 +\frac{2}{15}x^5 + \frac{17}{315}x^7+ \cdots}{x^3+\frac{1}{3}x^5 +\frac{2}{15}x^7 + \frac{17}{315}x^9+ \cdots} \quad \quad \small \color{#3D99F6}{\text{Dividing up and down by }x^3} \\ & = \lim_{x \to 0} \frac{\frac{1}{3} +\frac{2}{15}x^2 + \frac{17}{315}x^5+ \cdots}{1+\frac{1}{3}x^2 +\frac{2}{15}x^4 + \frac{17}{315}x^6+ \cdots} \\\ & = \frac{1}{3} \approx \boxed{0.333} \end{aligned}$

Series expansion makes life really easy.

Anurag Pandey - 5 years ago

Yes, if you use l'Hôpital's rule, it will be very long.

Chew-Seong Cheong - 5 years ago

No. It's not. After L'hopital rule, multiply top and bottom by $\cos^2 x$ . Let $L$ be this limit, find $1/L$ instead.

Pi Han Goh - 5 years ago

@Pi Han Goh – I see. Thanks

Chew-Seong Cheong - 5 years ago

Composites Done, Now What?

The answer is 0.333.

1 solution

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