I am still struggling with this one. Can't figure it out...
$$\lim_{x\to 0} \frac{1-\cos x}{x^2}$$
I tried $\cos x = \sqrt{1 - \sin^2{x}}$
Asked
Active
Viewed 108 times
3
-
2Apply L'Hôpital's rule twice – Simon S Apr 28 '15 at 16:32
-
6You could also use $1 - \cos(x) = 2\sin^2(x/2)$ – Gautam Shenoy Apr 28 '15 at 16:33
-
5Or multiply top and bottom by $1 + \cos x$. – Simon S Apr 28 '15 at 16:33
-
2Or use the Taylor expansion: $1-\cos(x) = \dfrac {x^2}{2} + O(x^4)$ – Apr 28 '15 at 16:33
-
If you put in $\cos x= \sqrt{1-\sin^2 x}$, then the fraction becomes $(1-\sqrt{1-\sin^2 x})/x^2$. Multiply the numerator and denominator by $(1+\sqrt{1-\sin^2 x})$, apply difference of squares to simplify the product in the numerator, and then you can apply the $(\sin x)/x\to 1$ limit. – Oscar Lanzi Dec 04 '18 at 13:34
1 Answers
3
Hint: $$\frac{1-\cos(x)}{x^2}=\frac{(1-\cos(x))(1+\cos(x))}{x^2(1+\cos(x))}=.....$$
Dr. Sonnhard Graubner
- 97,058