<s>[INST] How can I find this limit without L'Hopital and is my current solution correct? Find $$\lim_{x\rightarrow0} \frac{x^4}{1-2\cos x+\cos^2 x}$$
My solution:
Apply L'Hopital
$$=\lim_{x\rightarrow0} \frac{4x^3}{2\sin x-\sin 2x}$$
Apply L'Hopital again
$$=\lim_{x\rightarrow0} \frac{12x^2}{2\cos x-2\cos 2x}$$
Apply L'Hopital again
$$=\lim_{x\rightarrow0} \frac{24x}{-2\sin x + 4\sin 2x}$$
$$=\lim_{x\rightarrow0} \frac{24}{-2\cos x + 8\cos 2x}=\frac{24}{-2+8}=4$$
Is this correct? How would I approach this problem if I didn't want to use L'Hopital?
 [/INST] Your result looks right.
Without L'Hospital, here is a hint: $1-2\cos x+\cos^2 x=(1-\cos x)^2=4\sin^4\frac{x}{2}$
This follows from double-angle identity using half-angle: $\cos x=\cos^2\frac{x}{2}-\sin^2 \frac{x}{2}= 1-2\sin^2 \frac{x}{2} \implies 2\sin^2 \frac{x}{2}=1-\cos x$
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