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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

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writing \sin and \log instead of sin and log does not merely prevent italicization, but also results in proper spceing in things like $a\sin x$. I approved someone's edit and we're still waiting for a second approval. –  Michael Hardy Dec 18 '12 at 21:21

2 Answers 2

up vote 2 down vote accepted

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.

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the second one is pretty brutal with L'Hopital, although this may be what your teacher intended, I don't know. Anyway the answer winds up being $3/2$. –  user29743 Dec 18 '12 at 21:20
    
Could you give me the answers to the first and third one as well, so that I could check if my results are correct? –  Bilbo Dec 18 '12 at 21:29
    
@Anna: the solutions are (a) 1, (b) 3/2, and (c) $1/(10\ln 10)$. –  Fabian Dec 18 '12 at 22:26
    
Thanks, that's what I got ;) –  Bilbo Dec 18 '12 at 22:45

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.

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