How to solve $\lim\limits_{x\to 0} \frac{x - \tan(x)}{x^2}$ Without L'Hospital's Rule? you can use trigonometric identities and inequalities, but you can't use series or more advanced stuff.
Asked
Active
Viewed 218 times
1
-
1See http://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lhospital-rule-or-series-expansion – lab bhattacharjee Jan 12 '15 at 06:45
-
1The set up is a little off, it would be easy to show it is $0$. – Bombyx mori Jan 12 '15 at 06:54
-
Did you mean $x^3$ in the denominator (which is much more interesting than $x^2$)? In that case, it's a duplicate of http://math.stackexchange.com/q/157903/1242. – Hans Lundmark Jan 12 '15 at 07:27
-
Also very similar: http://math.stackexchange.com/q/134051/1242 – Hans Lundmark Jan 12 '15 at 09:59
1 Answers
1
Note that,
$\frac{\sin(x)-\tan(x)}{x^2} \leq \frac{x-\tan(x)}{x^2}\leq \frac{\tan(x)-\tan(x)}{x^2}=0$, for $\pi/2>x>0$ as $\sin(x) \leq x \leq \tan(x)$ for this interval.
For $x<0$ the inequalities are reversed because the functions tan and sin are even functions $\frac{\sin(x)-\tan(x)}{x^2} \geq \frac{x-\tan(x)}{x^2}\geq \frac{\tan(x)-\tan(x)}{x^2}=0$, for $-\pi/2<x<0$ as $\sin(x) \geq x \geq \tan(x)$ for this interval.
But, $\lim_{x \to 0} \frac{\sin(x)-\tan(x)}{x^2}=\lim_{x \to 0} \frac{\sin(x)}{x} \frac{\cos(x)-1}{x \cos(x)}=\lim_{x \to 0} \frac{-sin^2(x/2)}{(x/2)\cos(x)}=0$
So, $ 0 \leq \lim_{x \to 0} \frac{x-\tan(x)}{x^2} \leq 0$
Therefore, $\lim_{x \to 0} \frac{x-\tan(x)}{x^2} = 0$
Curious
- 1,278
-
You can't apply $\sin x\sim x$ in the second equality. – Martín-Blas Pérez Pinilla Jan 12 '15 at 08:36
-