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I did the following problem:

Find $\displaystyle\lim_{x\to 0} \frac{\cos^2x-1}{x^2}$

The following solution was given:

$$\lim_{x\to 0} \frac{\cos^2x-1}{x^2} = \lim_{x\to 0} \frac{2 \sin x \cdot \cos x}{2x} = 1$$

My questions regarding this solution are these:

In the denominator of the second limit in the solution above we have $2x$ and since we have $x$ aproaching $0$ this would mean division by $0$ which is not allowed. So, using L'Hôpital's rule, we should find the next higher derivative.

And regarding the numerator of this second limit I think, it should be $2 \cos x \cdot (- \sin x)$ by usage of the chain rule.

Thus I came the following solution:

$$\lim_{x\to 0} \frac{\cos^2x-1}{x^2} = \lim_{x\to 0} \frac{2 \cos x \cdot (- \sin x)}{2x} = \lim_{x\to 0} \frac{2\sin^2 x - 2 \cos^2 x}{2} = \frac{-2}{2} = -1$$

So, since these are different solutions, which is correct and where did I make a mistake? I would be thankful for explanations.

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    $\begingroup$ The expression inside the limit is nonpositive, so the limit cannot be $1$. $\endgroup$
    – L. F.
    Commented May 10, 2022 at 20:01
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    $\begingroup$ You are correct that the given solution has a sign error. As for invoking L'Hôpital again: that's certainly an option (and one you performed correctly). However, the given solution's author may be expecting the reader to recognize and exploit the fundamental result $\lim_{x\to0}\frac{\sin x}{x}=1$. ... Then again, if that were the case, a simpler solution would be to write $$\lim_{x\to0}-\frac{\sin^2x}{x^2}=-\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2=-\left(\lim_{x\to0}\frac{\sin x}{x}\right)^2=-1$$ $\endgroup$
    – Blue
    Commented May 10, 2022 at 20:10
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    $\begingroup$ You made a mistake but forgetting to put the negative in front of the 2 since derivative of $\cos x$ is $-\sin x$. Your second solution is correct. Also, you need to differentiate one more time which I see that you did the second time around. Using L'Hopital. $\endgroup$
    – homosapien
    Commented Jun 24, 2022 at 4:24
  • $\begingroup$ If my answer suits you, could you please mark it as correct? $\endgroup$
    – homosapien
    Commented Jun 27, 2022 at 0:10

1 Answer 1

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You have

$$\lim_{x \rightarrow 0} \frac{\cos^2x-1}{x^2}$$

Applying L'Hopital once you obtain

$$\lim_{x \rightarrow 0}\frac{-2\cos x \sin x}{2x}$$

this gives $0/0$ so apply L'Hopital once again to obtain

$$\lim_{x \rightarrow 0} \frac{-2(\cos^2x-\sin^2x)}{2}$$

sending $x$ to $0$ we get $-2/2=-1$ as needed.

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