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

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.
