Question

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

Answer 1

-2

Answer 2

Use L'Hopital's Rule

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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 \[\cos(0)=1 \]

Answer 7

Putting it all together your limit evaluates to -2.

Answer 8

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

Answer 9

Wonderful Manifold

Answer 10

\[\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1\]

is one limit that should be memorized. To see why it is true just look at any undergradute text on calculus.

Answer 11

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

Answer 12

It is true by squeeze thm or sandwich thm

Answer 13

whatever you want to call it

Answer 14

I usually go by squeeze theorem myself :)

Answer 15

i like sandwiches though

Answer 16

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

Answer 17

maybe that is wise