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UPSC NDA Math Model Paper 6

Question : 111 of 120

Marks: +1, -0

\lim\limits_{x\to 0} \frac{\cos 4x - \cos 6x}{x^2}

Solution:

Here, we have to find the value of

\lim\limits_{x\to 0} \frac{\cos 4x - \cos 6x}{x^2}

As we know that,

\cos C - \cos D = 2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{D-C}{2}\right)

So,

\cos 4x - \cos 6x = 2 \sin\left(\frac{4x+6x}{2}\right)

\sin\left(\frac{6x-4x}{2}\right)

= 2\sin 5x \sin x

Now,

\lim\limits_{x\to 0} \frac{\cos 4x - \cos 6x}{x^2}

= \lim\limits_{x\to 0} \frac{2\sin 5x \sin x}{x^2}

=2 \lim\limits_{x\to 0} \frac{\sin 5x}{x} \times \lim\limits_{x\to 0} \frac{\sin x}{x}

=2 \lim\limits_{x\to 0} \frac{\sin 5x}{5x} \times 5 \times 1

= 2 × 1 × 5

= 10

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