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anonymous

  • 5 years ago

find the limits when h ->0 h /tan^2h

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  1. anonymous
    • 5 years ago
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    h approaches 0 and tan^2h approaches 0 as h approaches 0, so you have an indeterminate form. Use L'Hopital's rule to obtain\[\lim_{h \rightarrow 0^+}\frac{h}{\tan^2h}=\frac{\lim_{h \rightarrow 0^+}h}{\lim_{h \rightarrow 0^+}\tan^2h}=\frac{\lim_{h \rightarrow 0^+}1}{\lim_{h \rightarrow 0^+}2\tan (h) \sec^2 h}=\frac{1}{\lim_{h \rightarrow 0^+}2\tan (h) \sec^2 h}\rightarrow +\infty\]

  2. anonymous
    • 5 years ago
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    Refresh the screen if it doesn't show up properly.

  3. anonymous
    • 5 years ago
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    where did u get sec

  4. anonymous
    • 5 years ago
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    That's the limit from the right (i.e. heading toward 0 from the positive part of the number line).

  5. anonymous
    • 5 years ago
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    From the derivative of tan(x):\[\frac{d}{dx}\tan^2x=2\tan x \sec ^2 x\](chain rule)

  6. anonymous
    • 5 years ago
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    i didnt know u take the derivative

  7. anonymous
    • 5 years ago
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    \[\lim_{h \rightarrow 0^-}\frac{h}{\tan^2h}=\frac{\lim_{h \rightarrow 0^-}h}{\lim_{h \rightarrow 0^-}\tan^2h}=\frac{\lim_{h \rightarrow 0^-}1}{\lim_{h \rightarrow 0^-}2\tan (h) \sec^2 h}=\frac{1}{\lim_{h \rightarrow 0^-}2\tan (h) \sec^2 h}\rightarrow -\infty\]

  8. anonymous
    • 5 years ago
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    You take the derivative for L'Hopital's rule.

  9. anonymous
    • 5 years ago
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    In its original form, you can't say anything about the limit. Anything where you have\[\frac{0}{0}, \pm \frac{\infty}{\infty}\]is an indeterminate form (you can't say anything about it). We use (usually) L'Hopital's rule to reduce it into a determinable form.

  10. anonymous
    • 5 years ago
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    Here: http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

  11. anonymous
    • 5 years ago
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    ohhhh,, nice

  12. anonymous
    • 5 years ago
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    so when u do that?

  13. anonymous
    • 5 years ago
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    when u cant do anything about the limit

  14. anonymous
    • 5 years ago
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    If you look above, I tell you when you have an indeterminate form.

  15. anonymous
    • 5 years ago
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    0/0 +/ infty/infty

  16. anonymous
    • 5 years ago
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    ok

  17. anonymous
    • 5 years ago
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    Here, the h in the numerator approached 0 the tan^2h in the denominator approached 0 when h approached 0, so you had a case of 0/0 which is indeterminate.

  18. anonymous
    • 5 years ago
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    You look at the numerator and denominator separately first (i.e. to see if you get an indeterminable form). If it's the case, apply L'Hopital's rule or something else (but no one uses the 'something else').

  19. anonymous
    • 5 years ago
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    jajjaaj

  20. anonymous
    • 5 years ago
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    ok

  21. anonymous
    • 5 years ago
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    :)

  22. anonymous
    • 5 years ago
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    thanks.. very good explanation

  23. anonymous
    • 5 years ago
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    welcome

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