First slide

Introduction to limits

Question

$If f (x) = |\begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1 \end{array}| then {Lt}_{x \to 0} \frac{f^{'} (x)}{x} =$

Moderate

A

0
B

2
C

-2
D

1

Solution

$f (x) = x |\begin{matrix} \cos x & 1 & 0 \\ 2 \sin x & x & x \\ \tan x & 1 & 0 \end{matrix}| = x (- x) (\cos x - \tan x) = x^{2} \tan x - x^{2} \cos x$
$\underset{x \to 0}{L t} \frac{f^{'} (x)}{x} = \underset{x \to 0}{L t} \frac{2 x \tan x + x^{2} \sec^{2} x - 2 x \cos x + x^{2} \sin x}{x}$
$= \underset{x \to 0}{L t} [2 \tan x + x \sec^{2} x - 2 \cos x + x \sin x] = - 2.$

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