# What is limit of: $\displaystyle\lim_{x\to 0}$$\tan x - \sin x\over x I want to search limit of this trigonometric function:$$\displaystyle\lim_{x\to 0}\frac{\tan x - \sin x}{x^n}$$Note: n \geq 1 - What do you know? What have you tried? Do you know what the Maclaurin expansion is? – Calvin Lin Jan 16 '13 at 10:58 If you know how to express \tan in terms of \sin and \cos, and you know something about the limits of (\sin x)/x^j and (1-\cos x)/x^k, you can put it all together to solve the problem. – Gerry Myerson Jan 16 '13 at 11:00 @CalvinLin I don't know it (Maclaurin expansion). I'm stuck in \lim_{x\to 0}$$(\sin x (1 - \cos x))\over x^n \cos x$. – lambda23 Jan 16 '13 at 11:06

Checking separatedly the cases for $\,n=1,2,3\,$, we find:

$$n=1:\;\;\;\;\;\;\frac{\tan x-\sin x}{x}=\frac{1}{\cos x}\frac{\sin x}{x}(1-\cos x)\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0$$

$$n=2:\;\;\frac{\tan x-\sin x}{x^2}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x}\xrightarrow [x\to 0]{}1\cdot 1\cdot 0=0\;\;(\text{Applying L'Hospital})$$

$$n=3:\;\;\;\frac{\tan x-\sin x}{x^3}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}\xrightarrow[x\to 0]{}1\cdot 1\cdot \frac{1}{2}=\frac{1}{2}\;\;(\text{Again L'H})$$

$$n\geq 4:\;\;\;\;\frac{\tan x-\sin x}{x^4}=\frac{1}{\cos x}\frac{\sin x}{x}\frac{1-\cos x}{x^2}\frac{1}{x^{n-3}}\xrightarrow[x\to 0]{}1\cdot 1\cdot \frac{1}{2}\cdot\frac{1}{\pm 0}=$$

and the above either doesn't exists (if $\,n-3\,$ is odd), or it is $\,\infty\,$ , so in any case $\,n\geq 4\,$ the limit doesn't exist in a finite form.

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If think that the limit exists for $n=2$ (limit is $0$) and for $n=3$ (limit is $\frac{1}{2}$). – Siméon Jan 16 '13 at 11:38
You're right, I didn't take into account the fact that we have a limit of the form $\,0/0\,$ in these cases. – DonAntonio Jan 16 '13 at 11:42

You can write this as a product $$\frac{\tan x -\sin x}{x^n} = \frac{\sin x(1-\cos x)}{x^n\cos x} = \frac{\sin x}{x}\times \frac{1-\cos x}{x^2} \times \frac{1}{\cos x}\times x^{3-n}$$ and you should be able to take the limits of the first three term as $x \to 0$.

Several behaviours are possible for $x^{3-n}$ as $x\to 0$ ,depending on the values of $n$.

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By Taylor expansion $$L=\lim_{x\to0}\frac{x^3/2+O(x^5)}{x^3} \cdot x^{(3 -n)}=\lim_{x\to0}\frac{x^{(3 -n)}}{2}$$ So, for $n=1,2 \rightarrow L=0$; $n=3 \rightarrow L=1/2$, $n$-even $\ge 4$ $\rightarrow$ L doesn't exist,$n$-odd $\ge 5$ $\rightarrow L=\infty$

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well to find the limit of [(tan(x)-sin(x))/x] as x approaches 0 just rewrite tan(x) as a quotient identity i.e tan(x) = sin(x)/cos(x) and divide each term by x separately then you will get sin(x)/(x cos(x))- sin(x)/x now we are free to use difference rule of limits because limit of sin(x)/(x*cos(x)) and sin(x)/x exist and finite i.e both 1. then limit of [(tan(x)-sin(x))/x] as x approaches 0 is limit of sin(x)/(x*cos(x)) as x approaches 0 minus limit of sin(x)/x as x approaches 0 which is 1-1=0. thnxs!

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