anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
-2
anonymous
  • anonymous
Use L'Hopital's Rule
anonymous
  • anonymous
\[ \lim_{x\rightarrow 0} \frac{x^2-2\sin(x)}{x} = \lim_{x\rightarrow 0} x - 2\cdot \lim_{x\rightarrow 0} \frac{\sin(x)}{x} \]

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

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