Okay, so we had this Math Game in class which, to avoid immersing you into irrelevant details, required the solution to the following limit:

$$\lim_{x \to 0}\frac{-x}{\tan(x)}$$

I got the right answer: -1, however, I was told the method I used to get the answer was not recommended.

This is what I did,

$$\lim_{x \to 0}\frac{-x}{\tan(x)} = \lim_{x \to 0}\frac{-x\cos (x)}{\sin(x)} = \frac{-1\cdot0 \cdot\cos(0)}{0} = -1\cdot\cos(0) = -1$$

The recommended path was--and still is:

$$\lim_{x \to 0}\frac{-x}{\tan(x)} = \lim_{x \to 0}\left[\frac{x}{\sin(x)} \cdot -\cos (x)\right]= 1\cdot-\cos(0) = -1$$

So, now I'm confused. Are we not essentially doing the same thing? The only difference is I'm canceling out the zeros by taking the limit of the expression first. Where as the correct process is taking the limit of $\frac{x}{\sin(x)}$, and then multiplying it by the limit of $\cos(x)$.

  • 3
    $\begingroup$ Cancelling zeros doesn't always work. Example: $$\lim_{x\to 0^+} \frac{x}{\left(\frac 1{\ln(x)}\right)} ``=" \frac{0}{0}$$ but the limit isn't $1$. $\endgroup$ – user137731 Sep 17 '16 at 21:45
  • $\begingroup$ Yes, you're essentially and directly doing the same...and it is right: you are not cancelling any zero (or at least you shouldn't...), but in fact using the very well known limit $$\lim_{x\to0}\frac{\sin x}x=\lim_{x\to0}\frac x{\sin x}=1$$ $\endgroup$ – DonAntonio Sep 17 '16 at 21:57
  • 6
    $\begingroup$ The second equality in your method isn't even a well-formed expression (what is $0/0$?), and the third one doesn't make any sense either. It happens to give the right answer in this particular case, but so does cancelling out the $6$s in $16/64 = 1/4$. A limit is not just mathematicians' jargon for substituting a number into a given expression. $\endgroup$ – anomaly Sep 17 '16 at 22:01

As you already know $$-\lim\limits_{x\to 0}\frac{x}{\tan x}=-\left(\lim\limits_{x\to 0}\frac{x}{\sin x}\right)\left(\lim\limits_{x\to 0}\cos x\right)$$ $$=-\lim\limits_{x\to 0}\cos x=-\cos 0 =-1$$ The above method clearly shows that the limit is well-defined as $-1$. Also note that a well-defined expression is not equal to an undefined expression. The problem with your method is that it introduces division by zero, which is ultimately undefined.


Besides noting that

$$\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{x}{\sin(x)} = 1$$

you could use LHR

$$\lim_{x \to 0} \frac{-x}{\tan(x)} = \lim_{x \to 0} \frac{-1}{\sec^2(x)} = -\lim_{x \to 0} \cos^2(x) = -1$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.