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.