6
$\begingroup$

Find the limit of $\lim\limits_{x\rightarrow0}\frac{x}{\tan x}$


Clearly, since the limit takes the form of $\frac{0}{0}$, one should try L'Hopital's Rule. If we apply L'Hopital's Rule, the problem is that $\frac{d(\tan x)}{dx}=\cot x$, and $\cot(0)$ is undefined. Thus, we cannot find a limit using L'Hopital's Rule.

My problem set suggests that this question can be answered using little more than L'Hopital and Cauchy's Mean Value Theorem. Not sure how to proceed. Please help.

$\endgroup$
  • 1
    $\begingroup$ $\frac{d}{dx} \tan x = \sec^2 x$. $\endgroup$ – William Stagner Mar 6 '15 at 3:52
  • $\begingroup$ should? ${}{}{}{}$ $\endgroup$ – hjhjhj57 Mar 6 '15 at 3:57
7
$\begingroup$

No, the derivative of $\tan x$ with respect to $x$ is $\sec^2x$, not $\cot x$. But you don’t need l’Hospital’s rule:

$$\frac{x}{\tan x}=\frac{x}{\sin x}\cdot\cos x\;,$$

and you should know the limits of both factors as $x\to 0$.

$\endgroup$
  • 1
    $\begingroup$ why should the limit of $\frac{\sin x}{x}$ be obvious/known and not $\frac{\tan x}{x}?$ $\endgroup$ – abel Mar 6 '15 at 4:15
  • 1
    $\begingroup$ @abel: Because it’s generally taught very early as a standard limit, at least in U.S. calculus courses, and $\lim_{x\to 0}\frac{\tan x}x$ isn’t. $\endgroup$ – Brian M. Scott Mar 6 '15 at 4:18
  • $\begingroup$ i see your reason but it seems odd to me that one is viewed more fundamental than the other. $\endgroup$ – abel Mar 6 '15 at 4:24
  • $\begingroup$ I have been lied to. This is why I consult the StackExchange. $\endgroup$ – Bob the Builds Mar 6 '15 at 14:31
5
$\begingroup$

Hint: $\lim_{x \to 0} \dfrac{x}{\sin x} = 1$ and $\dfrac{x}{\tan x} = \dfrac{x} {\sin x} \ \cos x$.

$\endgroup$
  • $\begingroup$ O mighty downvoter, can u explain why? $\endgroup$ – aNumosh Mar 6 '15 at 4:00
  • $\begingroup$ Might be the timing, since essentially the same answer had already been posted twice. The time gaps are very small, though, so that seems a bit harsh. I might not have otherwise, but in view of the downvotes I’m upvoting. $\endgroup$ – Brian M. Scott Mar 6 '15 at 4:16
4
$\begingroup$

Notice

$$ \frac{x}{ \tan x} = \frac{ \cos x}{\sin x} x \to 1 \times 1 = 1 $$ as $x \to 0 $ since $\frac{x }{\sin x} \to 1 $ as $x \to 0 $

$\endgroup$
  • $\begingroup$ False? ${}{}{}{}$ $\endgroup$ – Euler....IS_ALIVE Mar 6 '15 at 3:54
  • $\begingroup$ Whats wrong about it? This donwvotes definitely needs explanation $\endgroup$ – user203867 Mar 6 '15 at 3:57
  • 1
    $\begingroup$ It was wrong. Now it's correct. $\endgroup$ – Euler....IS_ALIVE Mar 6 '15 at 4:00
  • $\begingroup$ And the downvotes ought to be changed, but there’s no guarantee that they will be, unfortunately. It has my upvote, however. $\endgroup$ – Brian M. Scott Mar 6 '15 at 4:14
4
$\begingroup$

$\lim_{x\rightarrow 0} \frac{x} {\tan x}=\lim_{x\rightarrow 0} \frac{x\cos x} {\sin x}=\lim_{x\rightarrow 0} \frac{\cos x} {\frac{\sin x}{x}}=1$

$\endgroup$
  • $\begingroup$ Why downvote? anyone who downvote this please explain why? $\endgroup$ – Extremal Mar 6 '15 at 3:59
1
$\begingroup$

Here are the steps $$ \lim\limits_{x\to 0} \frac{x}{\tan x} = \lim\limits_{x\to 0} \frac{x}{\frac{\sin x}{\cos x}} = \lim\limits_{x\to 0} \frac{x\cos x}{\sin x}$$ $$ = \left(\lim\limits_{x\to 0} \frac{x}{\sin x}\right)\left(\lim\limits_{x\to 0} \cos x\right) = 1\cdot 1 = 1$$ Note that we can split the original limit into two convergent limits, as shown above.

$\endgroup$
0
$\begingroup$

Hope you know that $(\sin x)'=\cos x,(\cos x)'=\sin x$ and differentiating a fraction of functions. This will be very useful in your academic life: $$\frac{d}{dx}\tan x=\frac{d}{dx}\frac{\sin x}{\cos x}=\frac{\cos x.\cos x-\sin x.(-\sin x)}{\cos^2x}=\frac{\cos^2x+\sin^2x}{\cos^2x}=\sec^2x$$ So: $$\lim_{x\to0}\frac{x}{\tan x}=\lim_{x\to0}\frac{1}{\sec^2x}=\frac{1}{1^2}=1$$

$\endgroup$

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.