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?


1 Answer 1


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: $\cos x=\cos^2\frac{x}{2}-\sin^2 \frac{x}{2}= 1-2\sin^2 \frac{x}{2} \implies 2\sin^2 \frac{x}{2}=1-\cos x$

  • $\begingroup$ thank you! how do you know that $(1-\cos x)^2=4\sin^4(x/2)$ ? $\endgroup$ Oct 14, 2021 at 3:28
  • 1
    $\begingroup$ @user8290579 That is just a consequence of the half-angle identity $$\sin \frac{x}{2} = \sqrt{\frac{1 - \cos x}{2}}.$$ $\endgroup$
    – heropup
    Oct 14, 2021 at 3:29
  • 2
    $\begingroup$ +1. The Tireless Nitpicker (me) says the LHS in the half-angle formula in your comment should be encased in an Absolute-Value symbol. $\endgroup$ Oct 14, 2021 at 6:31
  • $\begingroup$ @user8290579: I edited the answer to clarify $\endgroup$
    – Vasili
    Oct 14, 2021 at 16:15

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