How to solve $\lim\limits_{x\to 0} \frac{x - \sin(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.
Solving $\lim\limits_{x\to0} \frac{x - \sin(x)}{x^2}$ without L'Hospital's Rule.
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0$\sin x \approx x-\frac{x^3}{6}$, you can read this http://nrich.maths.org/5622/solution and apply that to show this limit tends to $0$. – 2012-04-19
3 Answers
The given expression is odd; therefore it is enough to consider $x>0$. We then have 0<{x-\sin x\over x^2}<{\tan x -\sin x\over x^2}=\tan x\ {1-\cos x\over x^2}={\tan x\over2}\ \Bigl({\sin(x/2)\over x/2}\Bigr)^2\ , and right side obviously converges to $0$ when $x\to0+$.
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1First off, my point is that the answer above just converts one limit problem into another, without providing the solution for the second. It still does not. Second, how does that follow? x/2
for 0 , however ${x/2\over x}$ does not go to $1$. It goes to $1/2$. – 2012-04-20
We will in fact prove that $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} = \dfrac16$. This implies that $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^2} = 0$.
Let $S=\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3}$ Replacing $x$ by $2y$, we get that \begin{align} S & = \lim_{y \to 0} \dfrac{2y-\sin(2y)}{(2y)^3} = \lim_{y \to 0} \dfrac{2y-2 \sin(y) \cos(y)}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2y - 2 \sin(y) + 2 \sin(y) - 2 \sin(y) \cos(y)}{8y^3}\\ & = \lim_{y \to 0} \dfrac{2 y - 2 \sin(y)}{8y^3} + \lim_{y \to 0} \dfrac{2 \sin(y) - 2 \sin(y) \cos(y)}{8y^3}\\ & = \dfrac14 \lim_{y \to 0} \dfrac{y-\sin(y)}{y^3} + \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) (1 - \cos(y))}{y^3}\\ & = \dfrac{S}4 + \dfrac14 \lim_{y \to 0} \dfrac{\sin(y) 2 \sin^2(y/2)}{y^3}\\ & = \dfrac{S}4 + \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 + \dfrac18 \lim_{y \to 0} \dfrac{\sin(y)}{y} \lim_{y \to 0} \dfrac{\sin^2(y/2)}{(y/2)^2}\\ & = \dfrac{S}4 + \dfrac18\\ \dfrac{3S}4 & = \dfrac18\\ S & = \dfrac16 \end{align}
Hence, $\lim_{x \to 0} \dfrac{x-\sin(x)}{x^2} = \lim_{x \to 0} \left(\dfrac{x-\sin(x)}{x^3} \right)x = \left(\lim_{x \to 0} \dfrac{x-\sin(x)}{x^3} \right) \left( \lim_{x \to 0} x \right) = \dfrac{\lim_{x \to 0} x}6 = 0$
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4I might have missed something, but doesn't this answer show only that *if* $\lim\frac{x-\sin(x)}{x^3}$ exists *then* it is equal to $1/6$? – 2014-12-27
This can be done geometrically.
Surprisingly, two answers I wrote in this regard(geometric proofs of limits) before can be combined to give a solution for this.
$\lim_{x \to 0} \frac{ \tan x - x}{x^2} = 0 \tag{1}$
A geometric proof of that can be found here: Limit, solution in unusual way
$\lim_{x \to 0} \frac{1 - \cos x}{x} = 0 \tag{2}$
A geometric proof of that can be found here: Finding the limit of $(1-\cos(x))/x$ as $x\to 0$ with squeeze theorem
To combine the two:
$\tan x - x = \frac{\sin x - x \cos x}{\cos x} = \frac{(\sin x - x) + x(1 - \cos x)}{\cos x}$
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3Of course, you also need the geometric proof that $\lim_{x\to 0} \frac{\sin x}{x} = 1$. – 2012-04-19