## anonymous 4 years ago lim x→0 (x^2− 2 sin x)/x What do? I'm not a big fan of limits, so all help is appreciated.

1. anonymous

-2

2. anonymous

Use L'Hopital's Rule

3. anonymous

$\lim_{x\rightarrow 0} \frac{x^2-2\sin(x)}{x} = \lim_{x\rightarrow 0} x - 2\cdot \lim_{x\rightarrow 0} \frac{\sin(x)}{x}$

4. anonymous

Now the first of the two limits on the right evaluates to 0, clearly.

5. anonymous

And for the second, you can use L'Hopitals Rule or the Taylor expansion of sine, as you wish.

6. anonymous

By L'Hopitals Rule you get $\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = \lim_{x\rightarrow 0} \frac{(\sin(x))^\prime}{(x)^\prime} = \lim_{x\rightarrow 0} \frac{\cos(x)}{1} = 1$ where the ' denotes derivative with respect to x and the last equality holds because $\cos(0)=1$

7. anonymous

Putting it all together your limit evaluates to -2.

8. anonymous

Thank you that actually clears a lot of this limit thing up.

9. anonymous

Wonderful Manifold

10. Zarkon

$\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1$ is one limit that should be memorized. To see why it is true just look at any undergradute text on calculus.

11. myininaya

$\lim_{x \rightarrow 0}(\frac{x^2}{x}-2 \cdot \frac{\sin(x)}{x})=\lim_{x \rightarrow 0}x-2 \lim_{x \rightarrow 0}\frac{\sin(x)}{x}$ $0-2(1)=-2$

12. myininaya

It is true by squeeze thm or sandwich thm

13. myininaya

whatever you want to call it

14. Zarkon

I usually go by squeeze theorem myself :)

15. myininaya

i like sandwiches though

16. Zarkon

sandwiches are good...but i don't want to be thinking about food when I'm doing math

17. myininaya

maybe that is wise