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MarcLeclair

  • 2 years ago

Use L'hospital rule to find the following limit: Lim x-->0+ 6(tan(4x))^x I did this: Lim x-->0+ xln 6(tan(4x)) which is the same as: ln 6(tan(4x)) / (1/x) Take the derivative: (1/6tan(4x))) * 6(sec^2(4x))(4) / (-1/x^2). But I can't find a way to cancel out my denominator :/ the answer isn't infinity either

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  1. satellite73
    • 2 years ago
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    leave the six out of it, put it in at the end

  2. satellite73
    • 2 years ago
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    \[x\ln(\tan(4x)\] \[\frac{\ln(\tan(4x))}{\frac{1}{x}}\]etc

  3. satellite73
    • 2 years ago
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    \[-4x^2\csc(4x)\sec(4x)\] i think is the next step

  4. satellite73
    • 2 years ago
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    oh i see still a problem

  5. satellite73
    • 2 years ago
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    maybe flip the other way

  6. sirm3d
    • 2 years ago
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    \[\large= \frac{ 4x }{ \sin 4x }\frac{ -x }{ \cos 4x }\] there is no problem

  7. MarcLeclair
    • 2 years ago
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    your sin will make the denominator 0 even if multiplied :/

  8. MarcLeclair
    • 2 years ago
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    or am I just not understanding something ?

  9. satellite73
    • 2 years ago
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    no @sirm3d as it! first limit is 1

  10. satellite73
    • 2 years ago
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    *has it

  11. MarcLeclair
    • 2 years ago
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    0ooh take the limit of each... stupid me.

  12. MarcLeclair
    • 2 years ago
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    Thanks to both of you

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spraguer (Moderator)
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