# Determine the limit of $\lim_{x \rightarrow 0} \frac{\left | x \right | \cdot \left | \cos x \right |}{\left | \sin x \right |}$

Determine the limit of $$\lim_{x \rightarrow 0} \frac{\left | x \right | \cdot \left | \cos(x) \right |}{\left | \sin(x) \right |}$$

This is a task from an old exam. But I don't know if you can determine the limit of it at all because $sin$ and $cos$ oscillate.

If I just look at it, I cannot really say the limit. So I tried to use L'Hôpital's rule (derivate the enumerator and denominator):

$$\lim_{x \rightarrow 0} \frac{\cos(x)-x \cdot \sin(x)}{\cos(x)} = \lim_{x \rightarrow 0}\frac{\cos(x)}{\cos(x)}- \frac{x \cdot \sin(x)}{\cos(x)} = \lim_{x \rightarrow 0}\text{ }1 - \frac{x \cdot \sin(x)}{\cos(x)}$$

At this point I realized I have totally ignored the modulus signs :o

I'm not sure how to deal with them? Can I just set them after I finished derivating? So I would end up with:

$$\lim_{x \rightarrow 0}\text{ }1 - \frac{|x| \cdot |\sin(x)|}{|\cos(x)|}$$

Or just ignore the modulus signs and do the limit once going from left and once going from right side?

This is confusing but as it looks like in the end, it goes towards $1$ ?

• – R.W
Oct 9, 2017 at 21:12
• You can drop those modulus signs without any problem because the quantity inside those signs is positive. And use $x/\sin x\to 1$. Oct 10, 2017 at 10:06

Because $f(x) = |x|$ is continuous, you can exchange it with the limit, i.e. say that $$\lim_{x \to 0} \frac{|x| |\cos(x)|}{| \sin(x) |} = \lim_{x \to 0} \left| \frac{x \cos(x)}{\sin(x)}\right| = \left| \lim_{x \to 0} \frac{x \cos(x)}{\sin(x)}\right|$$ and the limit inside is easy to compute directly if you remember that $$\lim_{x \to 0} \frac{\sin x}{x} = 1.$$
• I do not like this argument too much. For instance $\lim\limits_{n\to\infty} 1=\lim\limits_{n\to\infty} |(-1)^n|=|\lim\limits_{n\to\infty} (-1)^n|$ which has no sense since $(-1)^n$ has no limit. I would prefer you say $\lim\limits_{x\to 0}\dfrac{x\cos(x)}{\sin(x)}=1$ so $x\mapsto|x|$ being continuous then $\lim\limits_{x\to 0}\left|\dfrac{x\cos(x)}{\sin(x)}\right|=1$. Your proof somehow took the reverse way.