Limits of three functions Could you help me calculate the following limits:
$$\lim_{x \to 0} x \left[  \frac{1}{x} \right]$$
$$\lim_{x\to 0}  \frac{1-\cos x \cdot \sqrt{\cos2x} }{x^2}$$
$$\lim_{x\to 10}  \frac{\log _{10}(x) - 1}{x-10}$$
As to the last one I thought I could use $\lim\frac{log _{a}(1+\alpha)}{\alpha} = \log_ae$ but it wouldn't work
 A: Here is a simpler way to calculate the second limit:
$$\lim_{x\to 0}  \frac{1-\cos x \sqrt{\cos2x} }{x^2}=\lim_{x\to 0}  \frac{1-\cos x \sqrt{\cos2x} }{x^2} \frac{1+\cos x \sqrt{\cos2x} }{1+\cos x \sqrt{\cos2x}}$$
$$=\lim_{x\to 0}  \frac{1-\cos^2 x \cos2x }{x^2} \lim_{x \to 0} \frac{1}{1+\cos x \sqrt{\cos2x}}$$
The second limit is easy, as for the first:
$$\lim_{x\to 0}  \frac{1-\cos^2 x \cos2x }{x^2}=\lim_{x\to 0}  \frac{1-\cos(2x)+\cos(2x)-\cos(2x)\cos^2 x }{x^2}$$
$$= \lim_{x \to 0} \frac{1-\cos(2x)}{x^2}+\lim_{x \to 0}\cos(2x) \frac{1-\cos^2 x }{x^2}$$
Now, use $1-\cos(2x)=2 \sin^2(x)$ and $1-\cos^2 x =\sin^2(x)$ and you are done.
A: Hints:
For the first limit, use the squeeze theorem - you know $\lfloor \frac{1}{x} \rfloor$ is trapped between $\frac{1}{x} - 1$ and $\frac{1}{x}$, and both of the corresponding limits are easy to compute.
For the second limit, you can use L'Hopital's rule twice (although this is gross).
The third limit is the derivative of the common logarithm at $x = 10$, which you can compute by expressing that logarithm in terms of the natural logarithm.
