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

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

Mathematics
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-2
Use L'Hopital's Rule
\[ \lim_{x\rightarrow 0} \frac{x^2-2\sin(x)}{x} = \lim_{x\rightarrow 0} x - 2\cdot \lim_{x\rightarrow 0} \frac{\sin(x)}{x} \]

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Now the first of the two limits on the right evaluates to 0, clearly.
And for the second, you can use L'Hopitals Rule or the Taylor expansion of sine, as you wish.
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 \[\cos(0)=1 \]
Putting it all together your limit evaluates to -2.
Thank you that actually clears a lot of this limit thing up.
Wonderful Manifold
\[\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1\] is one limit that should be memorized. To see why it is true just look at any undergradute text on calculus.
\[\lim_{x \rightarrow 0}(\frac{x^2}{x}-2 \cdot \frac{\sin(x)}{x})=\lim_{x \rightarrow 0}x-2 \lim_{x \rightarrow 0}\frac{\sin(x)}{x}\] \[0-2(1)=-2\]
It is true by squeeze thm or sandwich thm
whatever you want to call it
I usually go by squeeze theorem myself :)
i like sandwiches though
sandwiches are good...but i don't want to be thinking about food when I'm doing math
maybe that is wise

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