I am trying to figure out how to solve a certain set of problems using the trigonometric identities for evaluating limits, which are:

  • $\displaystyle \lim_{\theta \to 0} \frac{\sin\theta}{\theta}=1$

  • $\displaystyle \lim_{\theta\to 0} \frac{\cos\theta-1}{\theta}=0$

How can I apply these identities to the following problems? I am confused as to the actual calculus used to solve them. Also, I have just started calculus and I do not know any other rules to solve them with, such as L'Hopital's rule, which I have only heard the name of, and I have not gotten to that point in the class yet.

  1. $\displaystyle \lim_{x\to 0} \frac{\sin^2x}{x}$
  2. $\displaystyle \lim_{x\to 0} \frac{\tan^2x}{x}$
  3. $\displaystyle \lim_{x \to \pi/2 } \frac{\cos x}{\cot x}$
  4. $\displaystyle \lim_{x\to \pi /4} \frac{1-\tan x}{\sin x-\cos x}$
  5. $\displaystyle \lim_{t\to 0} \frac{\sin 3t}{2t}$


  • 1
    $\begingroup$ If $\lim_{x \to 0} f(x) = a$ and $\lim_{x \to 0} g(x) = b$, then $\lim_{x \to 0} f(x)g(x) = ab$. So, for example, for 1., take $f(x) = \frac{\sin x}{x}$ and $g(x) = \sin x$. $\endgroup$ – copper.hat Oct 3 '13 at 15:39
  • 2
    $\begingroup$ "these identities"...what identities are you talking about? $\endgroup$ – DonAntonio Oct 3 '13 at 15:41
  • $\begingroup$ @copper.hat I'm still a little confused by how that works out for #1. How could you solve for the limit with that information? $\endgroup$ – Justin W. Flory Oct 3 '13 at 15:48
  • $\begingroup$ @DonAntonio I'm referring to the special trigonometric limits (I think that's what they're called). I put them at the top of the question. $\endgroup$ – Justin W. Flory Oct 3 '13 at 15:48
  • $\begingroup$ $\frac{\sin^2 x}{x} = \sin x \frac{\sin x}{x}$. You know $\frac{\sin x}{x} \to 1$ and $\sin x \to 0$, hence $\frac{\sin^2 x}{x} \to 1\cdot 0 = 0$. $\endgroup$ – copper.hat Oct 3 '13 at 15:51

You should also know that, if $s(x)$ converges to $s$, and $t(x)$ converges to $t$, then $s(x)t(x)$ converges to $st$. Similarly, if $s \neq 0$ and $s(x) \neq 0$ for all $x$ (you can restrict attention to $x < \epsilon$) then $t_n/s_n$ converges to $t/s$.

Given this, the first limit can be viewed as

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

  • 1
    $\begingroup$ I'm also confused, since the example given is wrong. Also, you have a typo, $t_n/s_n$ converges to $t/s$, provided $s\neq 0$. $\endgroup$ – Jean-Sébastien Oct 3 '13 at 15:53
  • $\begingroup$ Thanks for catching the typo. Which example are you referring to? $\endgroup$ – BaronVT Oct 3 '13 at 16:16
  • $\begingroup$ $\sin(x)/x$ converges to $1$ as $x$ goes to $0$. $\endgroup$ – Jean-Sébastien Oct 3 '13 at 17:01

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