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Limits Continuity and Differentiability

Section: Mathematics

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Question : 59 of 61

Marks: +1, -0

\lim\limits_{x\to 0} \frac{1-\cos x \cos 2x}{\sin^2 x}=

[AP EAPCET 18 May 2024 Shift 1]

$\frac{11}{4}$
$\frac{5}{2}$
3
5

Solution:

Step 1 - Expand the numerator using small-angle approximations

As

x\to 0

:

\cos x = 1-\frac{x^2}{2}+O(x^4),\qquad \cos 2x = 1-2x^2+O(x^4)

Multiply:

\cos x \cos 2x = \left(1-\frac{x^2}{2}\right)(1-2x^2)+O(x^4)

= 1-2x^2-\frac{x^2}{2}+O(x^4)

= 1-\frac{5}{2}x^2+O(x^4)

Thus:

1-\cos x \cos 2x = \frac{5}{2}x^2+O(x^4)

Step 2 - Expand the denominator

\sin x = x+O(x^3) \quad \Rightarrow \quad \sin^2 x = x^2+O(x^4)

Step 3 - Form the ratio

\frac{1-\cos x \cos 2x}{\sin^2 x} = \frac{\frac{5}{2}x^2+O(x^4)}{x^2+O(x^4)} \to \frac{5}{2}

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