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Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)

$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$ since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,

$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$

since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,

$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$

since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.

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    $\begingroup$ Minor correction: the derivative of $\sec^2x-1$ is $2\sec^2x\tan x$, not $2\sec^2x\sec x\tan x$. (This doesn't affect the limit, since $\sec0=1$.) The key mistake is saying that the quotient always goes to zero [divided] by zero. One more round of L'Hopital would give $-\cos0=-1$ in the numerator. $\endgroup$ – Barry Cipra Apr 21 '16 at 13:27
  • $\begingroup$ @BarryCipra : But in this case there's a simpler thing to do than "one more round with L'Hopital", as in the answer posted by user49685. $\qquad$ $\endgroup$ – Michael Hardy Apr 21 '16 at 13:58
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    $\begingroup$ $\tan x$ in denominator is $\sin x/\cos x$ so if you multiply the last fraction by $\cot x/\cot x$ you'll get $$\frac{-\cos x}{2\sec^3 x}$$ $\endgroup$ – CiaPan Apr 21 '16 at 14:02
  • $\begingroup$ @MichaelHardy, agreed. Even the second round of L'Hopital was unnecessary, as randomgirl's answer (implicitly) shows. I was just addressing the OP's immediate misunderstanding. $\endgroup$ – Barry Cipra Apr 21 '16 at 14:10
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$$\frac{\cos(x)-1}{\sec^2(x)-1}=\frac{\cos(x)-1}{\frac{1}{\cos^2(x)}-1}=\frac{\cos^2(x)(\cos(x)-1))}{1-\cos^2(x)}=\frac{\cos^2(x)(\cos(x)-1)}{(1-\cos(x))(1+\cos(x))}=\frac{- \cos^2(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))}$$

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Notice that $\tan x = \sin x \sec x$; so we'll have:

$$\lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec x \tan x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^2x\sec^2 x \sin x} = \lim_{x \to 0} \frac{-\sin x}{2\sec^4x\sin x} = \lim_{x \to 0} \frac{-1}{2\sec^4x} = \cdots$$

Hope you can go from here. :x

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For $x\to 0$ we have $\sin x\sim x-\frac{x^3}{3!} $ and $\tan x\sim x+\frac{x^3}{3} $ then $$ \frac{\sin x- x}{\tan x-x}\sim \frac{-\frac{x^3}{3!}}{+\frac{x^3}{3}}=-\frac{1}{2} $$

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Hint: use the limit development of sin and tan

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  • $\begingroup$ A far simpler thing to do is to realize that $\dfrac{\sin x}{\tan x} = \cos x$. $\qquad$ $\endgroup$ – Michael Hardy Apr 21 '16 at 13:54
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Hint differentiate again you get $$\frac{-\cos(x)}{2(\sec^4(x)+2\tan^2(x)\sec^2(x))}$$ now whats the limit

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$$ \lim_{x\to0}\frac{\sin x-x}{\tan x-x}\overset{L'H}{=}\lim_{x\to0}\frac{\cos x-1}{\frac{1}{\cos^2x}-1}=\lim_{x\to0}\frac{\frac{\cos x-1}{x^2}}{\frac{1-\cos^2 x}{x^2\cos^2x}} $$ now, $$ \lim_{x\to0}\frac{\cos x-1}{x^2}=-\frac{1}{2} $$ and $$ \lim_{x\to0}\frac{1-\cos^2 x}{x^2\cos^2x}=\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\frac{1}{\cos^2x}=1 $$ so, the desired limit is $-\frac{1}{2}$

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You can make it a little simpler by rewriting

$$\frac{\sin(x)-x}{\tan(x)-x}=\cos(x)\frac{\sin(x)-x}{\sin(x)-x\cos(x)}.$$

The first factor tends to $1$ and can be ignored. Then, applying L'Hospital three times,

$$\frac{\sin(x)-x}{\sin(x)-x\cos(x)}\to\frac{\cos(x)-1}{\cos(x)-\cos(x)+x\sin(x)}\to\frac{-\sin(x)}{\sin(x)+x\cos(x)}\to\frac{-\cos(x)}{\cos(x)+\cos(x)-x\sin(x)}\to-\frac12.$$


Alternatively,

$$\frac{\cos(x)-1}{x\sin(x)}=\frac{-2\sin^2\left(\dfrac x2\right)}{4\left(\dfrac{x}2\right)^2}\frac x{\sin(x)}\to-\frac24\cdot1^2\cdot1.$$

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