Lazy Limits 5!

Calculus Level 3

$\large L = \displaystyle\large\lim_{x\to 0}\dfrac{\sin^{-1} x - \tan^{-1} x}{x^3}$

Find $L$ .

The answer is 0.5.

1 solution

Chew-Seong Cheong
Jul 12, 2016

$\begin{aligned} L & = \lim_{x \to 0} \frac {\color{#3D99F6}{\sin^{-1}x} - \color{#3D99F6}{\tan^{-1}x} }{x^3} \quad \quad \small \color{#3D99F6}{\text{By Maclaurin series (note that }|x| < 1)} \\ & = \lim_{x \to 0} \frac {\color{#3D99F6}{\left(x + \frac 16 x^3 + \frac 3{40} x^5+...\right)} - \color{#3D99F6}{\left(x - \frac 13 x^3 + \frac 15 x^5-...\right)}}{x^3} \\ & = \lim_{x \to 0} \frac {\frac 12 x^3 - \frac 18 x^5 + O(x^7)}{x^3} \quad \quad \small \color{#3D99F6}{\text{Divide up and down by }x^3} \\ & = \lim_{x \to 0} \frac {\frac 12 - \frac 18 x^2 + O(x^4)}{1} \\ & = \frac 12 = \boxed{0.5} \end{aligned}$

Sir , could you please explain that 'o' notation in brief.

Syed Shahabudeen - 4 years, 11 months ago

It is a short way to write $a_4x^4+a_6x^6 + a_8x^8 +...$ . That is $O(x^4) = a_4x^4+a_6x^6 + a_8x^8 +...$ which means that the following terms have $x$ with powers 4 and higher. So when $x \to 0$ , $O(0) = 0$ .

Chew-Seong Cheong - 4 years, 11 months ago

Lazy Limits 5!

The answer is 0.5.

1 solution

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