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I was solving the limit $$\lim_{x\to0}\frac{\sqrt{\frac{1}{\cos x}}-1}{\sin^2 {\frac{x}{16}}}$$ I simplified the numerator by multiplying both the numerator and denominator by $\sqrt{\frac{1}{\cos x}}+1$ getting to $$\frac{1-\cos x}{\sin^2 \left(\frac x {16}\right) \left(\sqrt{\frac{1}{\cos x}}+1 \right)\cos x}$$ After this step, Wolfram suggests to use the product rule and calculate separately: $$\lim_{x\to0}\frac{1-\cos x}{\sin^2\left(\frac{x}{16}\right)}\;\cdot\;\lim_{x\to0}\frac 1 {\left(\sqrt{\frac{1}{\cos x}}+1\right)\cos x}$$ which eventually gets to the final result of $64$ using De l'Hopital.

I was wondering if there was a better/simpler solution that doesn't require De l'Hopital as my professor suggests to avoid using that as much as possible.

Thanks in advance.

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    $\begingroup$ $\cos x=1-2\sin^2\frac{x}2$ would come in handy, as well as the limit $\frac{\sin \theta}{\theta}\to 1$. $\endgroup$
    – user10354138
    Sep 10 at 15:17

3 Answers 3

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By Taylor's formula, $$\sqrt{\sec x}=\sqrt{1+\frac{x^2}{2}+O(x^4)}=1+\frac{x^2}{4}+O(x^4)$$ by Taylor's formula, $$\sin^2\frac{x}{16}=\left(\frac{x}{16}+O(x^3)\right)^2=\frac{x^2}{256}+O(x^4)$$ both as $x\to0$. Thus, $$\frac{\sqrt{\sec x}-1}{\sin^2(x/16)}=\frac{x^2/4+O(x^4)}{x^2/256+O(x^4)}$$ as $x\to 0$. The limit is $64$.

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  • $\begingroup$ Sorry for the late comment, how is the first formula going from $\sqrt{1+\frac{x^2}2+o(x^4)}$ to $1+\frac{x^2}{4}+o(x^4)$? $\endgroup$
    – Green
    Sep 21 at 16:24
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    $\begingroup$ Use the binomial series and expand up to the $2$nd degree terms. $\endgroup$
    – bob
    Sep 21 at 17:44
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$$\lim_{x\to0}\frac{\sqrt{\frac{1}{\cos x}}-1}{\sin^2 {\frac{x}{16}}}=$$ $$=\lim_{x\to0}\frac{\sqrt{\frac{1}{\cos x}}-1}{\sin^2 {\frac{x}{16}}}\times \frac{\sqrt{\frac{1}{\cos x}}+1}{\sqrt{\frac{1}{\cos x}}+1}$$ $$=\lim_{x\to0}\frac{\frac{1}{\cos x}-1}{\sin^2 {\frac{x}{16}}\times (\frac{1}{\cos x}+1)}=$$ $$=\lim_{x\to0}\frac{\frac{1-\cos x}{\cos x}}{\sin^2 {\frac{x}{16}}\times (\frac{1}{\cos x}+1)}=$$ $$=\lim_{x\to0}\frac{\frac{1-(1-2\sin^2\frac{x}{2})}{\cos x}}{\sin^2 {\frac{x}{16}}\times (\frac{1}{\cos x}+1)}=$$ $$=\lim_{x\to0}\frac{\frac{2\sin^2\frac{x}{2}}{\cos x}}{\sin^2 {\frac{x}{16}}\times (\frac{1}{\cos x}+1)}=$$ $$=\lim_{x\to0}\frac{2\sin^2\frac{x}{2}}{\sin^2 {\frac{x}{16}}\times (\frac{1}{\cos x}+1)\times \cos x}=$$ $$=\frac{2\times (\frac{1}{2})^2}{ {(\frac{1}{16})^2}\times 2\times 1}=64$$

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  • $\begingroup$ Really good alternative explaination, i struggled in choosing one to accept, thanks! $\endgroup$
    – Green
    Sep 10 at 17:40
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Let $L$ denote the limit. Observe that

$$\frac{\sqrt{\frac{1}{\cos x}}-1}{\sin^2 {\frac{x}{16}}} =16^2\left(\frac{\frac{x}{16}}{\sin\frac{x}{16}}\right)^2\frac{1-\sqrt{\cos x}}{x^2}\frac{1}{\sqrt{\cos x}}.$$

By continuity, the standard limit

$$\lim_{\xi\to0}\frac{\sin\xi}{\xi}=1,$$

and the limit

$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}=\frac{1}{4}$$

from this answer, we get that

$$L=16^2\cdot 1\cdot \frac{1}{4}\cdot1=64.$$

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