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anonymous

  • 4 years ago

if f(x)=tanx-x and g(x)=x^3, evaluate the limit of f(x) over g(x) as x approaches 0. -3rd question.

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  1. anonymous
    • 4 years ago
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    @zarkon, et al, apparently she needs a non-l'hopital method, and i'll be darned if i can think of one

  2. Zarkon
    • 4 years ago
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    how about a power series solution :)

  3. anonymous
    • 4 years ago
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    yes, do not pass go, do not collect $200

  4. Zarkon
    • 4 years ago
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    lol

  5. Zarkon
    • 4 years ago
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    go directly to jail

  6. anonymous
    • 4 years ago
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    in fact i think you need l'hopital twice at least

  7. Zarkon
    • 4 years ago
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    power series gives the quickest solution

  8. anonymous
    • 4 years ago
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    \[\frac{\sec^2(x)-1}{3x^2}\] \[\frac{\tan^2(x)}{3x^2}\] etc

  9. anonymous
    • 4 years ago
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    yes but if this is pre-derivative, then what can you do?

  10. nenadmatematika
    • 4 years ago
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    kaezalorene you can always say to your teacher that you've discovered some rule when you were home studying.....the rule that you can use the derivatives and solve....he will be suprised :D

  11. nenadmatematika
    • 4 years ago
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    kidding :D

  12. ash2326
    • 4 years ago
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    \[f(x)=\frac{sin x}{cosx}-x\] \[g(x)=x^3\] we have \[f(x)/ g(x)=\frac{\frac{sin x}{cosx}-x}{x^3}\] dividing numerator and denominator by x now \[f(x)/g(x)= \frac{\frac{sin x}{x}-cos x}{x^2}\] \[limit x-->0 \frac{sinx}{x}=1\] so we have \[f(x)/g(x)= \frac{1-cos x}{x^2}\] 1-cos x = 2 (sin x/2)^2 so we get \[f(x)/g(x)= \frac{2 sin ^2 x/2}{x^2}\] dividing the numerator and denominator by 4 we get \[f(x)/g(x)=\frac {\frac{2 sin ^2 x/2}{4}}{\frac{x^2}{4}}\] \[(sin x / 2)/ (x/2-)-->1 as x---->0\] so we get \[f(x)/g(x)= \frac {2}{4}\] = 1/2

  13. ash2326
    • 4 years ago
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    I missed a cos x in the denominator but it won't make any difference as x-->0 cos x ---> 1

  14. ash2326
    • 4 years ago
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    Zarkon what do you think , is my method correct?

  15. Zarkon
    • 4 years ago
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    the answer is 1/3

  16. Zarkon
    • 4 years ago
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    unfortunatly you can't do it like that (the sin(x)/x is an integral part of the final solution...you can't takes its limit while holding everything else constant)

  17. ash2326
    • 4 years ago
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    Thanks I got your point

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