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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.

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    $\begingroup$ $\frac{d}{dx} \tan x = \sec^2 x$. $\endgroup$ Mar 6, 2015 at 3:52
  • $\begingroup$ should? ${}{}{}{}$ $\endgroup$
    – hjhjhj57
    Mar 6, 2015 at 3:57

6 Answers 6

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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 $x\to 0$.

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    $\begingroup$ why should the limit of $\frac{\sin x}{x}$ be obvious/known and not $\frac{\tan x}{x}?$ $\endgroup$
    – abel
    Mar 6, 2015 at 4:15
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    $\begingroup$ @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. $\endgroup$ Mar 6, 2015 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, 2015 at 4:24
  • $\begingroup$ I have been lied to. This is why I consult the StackExchange. $\endgroup$ Mar 6, 2015 at 14:31
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Hint: $\lim_{x \to 0} \dfrac{x}{\sin x} = 1$ and $\dfrac{x}{\tan x} = \dfrac{x} {\sin x} \ \cos x$.

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  • $\begingroup$ O mighty downvoter, can u explain why? $\endgroup$
    – anumosh
    Mar 6, 2015 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. $\endgroup$ Mar 6, 2015 at 4:16
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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 $

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  • $\begingroup$ False? ${}{}{}{}$ $\endgroup$ Mar 6, 2015 at 3:54
  • $\begingroup$ Whats wrong about it? This donwvotes definitely needs explanation $\endgroup$
    – user203867
    Mar 6, 2015 at 3:57
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    $\begingroup$ It was wrong. Now it's correct. $\endgroup$ Mar 6, 2015 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. $\endgroup$ Mar 6, 2015 at 4:14
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$\lim_{x\rightarrow 0} \frac{x} {\tan x}=\lim_{x\rightarrow 0} \frac{x\cos x} {\sin x}=\lim_{x\rightarrow 0} \frac{\cos x} {\frac{\sin x}{x}}=1$

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  • $\begingroup$ Why downvote? anyone who downvote this please explain why? $\endgroup$
    – Extremal
    Mar 6, 2015 at 3:59
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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.

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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: $$\lim_{x\to0}\frac{x}{\tan x}=\lim_{x\to0}\frac{1}{\sec^2x}=\frac{1}{1^2}=1$$

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