Question

compute: [\lim_{x\to 0}\frac{\sqrt{\cos(x)}-\cos(x)}{x^2}] without h'opital

Answer 1

You have to ask the smart people about this.

Answer 2

and whom they might be? :)

Answer 3

\[\lim_{x\to 0}\frac{\sqrt{\cos(x)}-\cos(x)}{x^2}\]
one idea I got was to try to factor out a sin^2 so u can make lim x to 0 of sin^2/x^2 go to 1

Answer 4

\[\lim_{x \rightarrow 0} \frac{\sqrt{\cos(x)} - \cos(x)}{x²} = \lim_{x \rightarrow 0} \frac{\sqrt{\cos(x)} \cdot (1 - \sqrt{\cos(x)})}{x²}\]\[=\lim_{x \rightarrow 0} \frac{\sqrt{\cos(x)}\cdot (1 - \cos(x))}{x²\cdot (1+\sqrt{\cos(x)})} = \lim_{x \rightarrow 0} \frac{\sqrt{\cos(x)}\cdot 2\sin²(\frac{x}{2})}{x² \cdot (1 + \sqrt{\cos(x)})}\]

Answer 5

\[ = \lim_{x \rightarrow 0} 2 \frac{\sin²(\frac{x}{2})}{x²\cdot (\frac{1}{\sqrt{\cos(x)}} + 1)}\]\[= \lim_{x \rightarrow 0} 2 \cdot \frac{\sin\frac{x}{2}}{x} \cdot \frac{\sin\frac{x}{2}}{x} \cdot \frac{1}{\frac{1}{\sqrt{\cos(x)}}+1}\]

Answer 6

Now I'm going to calculate the respective limits of each factor and thus show that they exist, so that we can use the multiplication property of limits to calculate the limit of their product. First off, I'll use l'Hospital's Rule (I hope you're familiar with it):
\[\lim_{x \rightarrow 0} \frac{\sin(\frac{x}{2})}{x} = \lim_{x \rightarrow 0} \cos(\frac{x}{2}) \cdot \frac{1}{2} = \frac{1}{2}\] \[\lim_{x \rightarrow 0} \frac{1}{\frac{1}{\sqrt{\cos(x)}}+1} = \frac{1}{2}\] It follows that our original limit is equal to \[2 \cdot (\frac{1}{2})^3 = \frac{1}{4}\]