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dgamma3

  • 3 years ago

How would you solve the following limit: limit (x -> 0+) (x-11)/sin(x)

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  1. dgamma3
    • 3 years ago
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    you can do it manually, and figure out its negative infinity. but is there any algebraic way.

  2. dgamma3
    • 3 years ago
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    \[\lim_{x \rightarrow 0+} (x-11)/\sin(x)\]

  3. nitz
    • 3 years ago
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    \[\lim_{x \rightarrow 0}(sinx/x)=1\]

  4. kirbykirby
    • 3 years ago
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    \[You know \lim_{x \rightarrow 0+} \frac{\sin x}{x}=1 \]

  5. kirbykirby
    • 3 years ago
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    so: do the same trick I told you before: Divide the top and bottom by x

  6. nitz
    • 3 years ago
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    and \[\lim_{x \rightarrow 0}(11/sinx)\rightarrow \infty \]

  7. kirbykirby
    • 3 years ago
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    \[\lim_{x \rightarrow 0+} \frac{\frac{11-x}{x}}{\frac{\sin x}{x}} = \lim_{x \rightarrow 0+} \frac{11/x-1}{\frac{\sin x}{x}}=\]\[\frac{\infty-1}{1}=\infty\]

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