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I need help finding the the following limit:

$$\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x} $$

I tried to simplify to:

$$ \lim_{x\rightarrow 0} \frac{\sin x \cos x}{x\cos x+\sin x} $$

but I don't know where to go from there. I think, at some point, you have to use the fact that $\lim_{x\rightarrow 0} \frac{\sin x}{x} = 1$. Any help would be appreciated.

Thanks!

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  • $\begingroup$ How about a Taylor-Series expansion? If you know $\lim\limits_{x\to0} \frac{\sin x}x = 1$, you'll know about the limits of polynomial quotients... $\endgroup$ – AlexR Dec 18 '13 at 22:09
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$$ \frac{\sin x}{x + \tan x} = \frac{1}{\frac{x}{\sin x}+\frac{\tan x}{\sin x}} \to 1/2 $$

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Turn it upside down. ${}{}{}{}{}{}$

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  • $\begingroup$ How can I just turn it upside down? $\endgroup$ – Jeel Shah Dec 18 '13 at 22:40
  • $\begingroup$ Take the reciprocal. We get $\frac{x}{\sin x}+\frac{1}{\cos x}$. $\endgroup$ – André Nicolas Dec 18 '13 at 22:47
  • $\begingroup$ I can do that? Don't we have take the reciprocal somewhere else to? I'm not understanding how the algebra works out. $\endgroup$ – Jeel Shah Dec 18 '13 at 22:59
  • $\begingroup$ We take the reciprocal, and show easily that it has limit $2$. So the original expression has limit $\frac{1}{2}$. $\endgroup$ – André Nicolas Dec 18 '13 at 23:01
  • $\begingroup$ Ahh, so originally it's something like $\lim_{x\to0} \frac{\sin x}{x + \tan x} = L$ and then I can reciprocal and it will make sense since, we also reciprocal the right side to $1\over L$ right? $\endgroup$ – Jeel Shah Dec 18 '13 at 23:04
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Hint

Divide both numerator and denominator by $x$ and use the fact $$\lim_{x\to0}\frac{\sin x}{x}=1$$

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$$\lim_{x\rightarrow 0} \frac{\sin x}{x + \tan x} =\lim_{x\rightarrow 0} \frac{1}{\frac{x}{\sin x} + \frac{1}{cos x}} =1/2$$

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$$\lim_{x\rightarrow 0}\frac{\sin x}{x+\tan x}=\lim_{x\rightarrow 0}\frac{\frac{\sin x}{x}}{1+\frac{\tan x}{x}}=\frac {1}{1+1}=\frac{1}{2}$$

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Given that $$ \lim_{x \to 0} \frac{\sin x}{x} = 1, $$ we can calculate that $$ \begin{align} \lim_{x \to 0} \frac{\tan x}{x} &= \lim_{x \to 0} \frac{\sin x}{x \cos x} \\ &= \lim_{x \to 0} \frac{\sin x} {x} \cdot \lim_{x \to 0} \frac{1}{\cos x} \\ &= 1. \end{align} $$


Now, rewrite your limit and compute: $$ \lim_{x \to 0} \frac{\sin x}{x + \tan x} = \lim_{x \to 0} \frac{\frac{\sin x}{x}}{1 + \frac{\tan x}{x}} = \frac{1}{2}. $$

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$$\sin x \sim x$$ $$\tan x \sim x$$

Hence

$$\frac{\sin x}{x + \tan x} \sim \frac{x}{x+x} = \frac{1}{2}$$

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Look at the ratio of the Taylor series: $\frac{\sin x}{x+\tan x} = \frac{x-\frac{x^3}{3!}+\cdots}{x + (x+\frac{x^3}{3}+\cdots)}$. Combine the $x$ terms in the denominator, divide out a factor of $x$, and you have $\frac{1 - \frac{x^2}{2!}+\cdots}{2 + \frac{x^2}{3} + \cdots}$. Then you can let $x\rightarrow 0$ to get $\frac12$ because only the constant terms remain in the limit.

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