First slide

Introduction to limits

Question

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

Moderate

A

1
B

-1
C

2
D

-2

Solution

$\begin{array}{l} f (x) = \cos x [x^{2} - 2 x^{2}] - x [2 \sin x - 2 x \tan x] + 1 [2 x \sin x - x^{2} \tan x] \\ = - x^{2} \cos x - 2 x \sin x + 2 x^{2} \tan x + 2 x \sin x - x^{2} \tan x \\ = - x^{2} \cos x + x^{2} \tan x \\ f^{1} (x) = - 2 x \cos x + x^{2} \sin x + 2 x \tan x + x^{2} \sec^{2} x \\ lim_{x \to 0} \frac{f^{1} (x)}{x} = lim_{x \to 0} \frac{- 2 x \cos x + x^{2} \sin x + 2 x \tan x + x^{2} \sec^{2} x}{x} \\ = lim_{x \to 0} - 2 \cos x + x \sin x + 2 \tan x + x \sec^{2} x \\ = - 2 + 0 + 0 + 0 = - 2 \end{array}$

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