I assume you meant:
$$\lim_{x \to 0} \frac{2\cos(x) + \cos(2x)}{x}$$
Now, let's use the double-angle formula for cosine, which is:
$$\cos(2x) = 2\cos^2(x) - 1$$
Substitute this into the expression:
$$\lim_{x \to 0} \frac{2\cos(x) + 2\cos^2(x) - 1}{x}$$
Now,
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