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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.

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  1. anonymous
    • 4 years ago
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    -2

  2. anonymous
    • 4 years ago
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    Use L'Hopital's Rule

  3. anonymous
    • 4 years ago
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    \[ \lim_{x\rightarrow 0} \frac{x^2-2\sin(x)}{x} = \lim_{x\rightarrow 0} x - 2\cdot \lim_{x\rightarrow 0} \frac{\sin(x)}{x} \]

  4. anonymous
    • 4 years ago
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    Now the first of the two limits on the right evaluates to 0, clearly.

  5. anonymous
    • 4 years ago
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    And for the second, you can use L'Hopitals Rule or the Taylor expansion of sine, as you wish.

  6. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    Putting it all together your limit evaluates to -2.

  8. anonymous
    • 4 years ago
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    Thank you that actually clears a lot of this limit thing up.

  9. anonymous
    • 4 years ago
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    Wonderful Manifold

  10. Zarkon
    • 4 years ago
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    \[\lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1\] is one limit that should be memorized. To see why it is true just look at any undergradute text on calculus.

  11. myininaya
    • 4 years ago
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    \[\lim_{x \rightarrow 0}(\frac{x^2}{x}-2 \cdot \frac{\sin(x)}{x})=\lim_{x \rightarrow 0}x-2 \lim_{x \rightarrow 0}\frac{\sin(x)}{x}\] \[0-2(1)=-2\]

  12. myininaya
    • 4 years ago
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    It is true by squeeze thm or sandwich thm

  13. myininaya
    • 4 years ago
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    whatever you want to call it

  14. Zarkon
    • 4 years ago
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    I usually go by squeeze theorem myself :)

  15. myininaya
    • 4 years ago
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    i like sandwiches though

  16. Zarkon
    • 4 years ago
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    sandwiches are good...but i don't want to be thinking about food when I'm doing math

  17. myininaya
    • 4 years ago
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    maybe that is wise

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