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Valdas

  • 3 years ago

Finding a limit

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  1. Valdas
    • 3 years ago
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    \[\lim_{x \rightarrow 0} \frac{ \sqrt[5]{1+2x}-1 }{ \sin x }\]

  2. myko
    • 3 years ago
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    try l`hopital

  3. Valdas
    • 3 years ago
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    sorry, I forgot to mention that we haven't learnt about l`hopital

  4. myko
    • 3 years ago
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    nvm

  5. myko
    • 3 years ago
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    you learned Taylors series?

  6. Valdas
    • 3 years ago
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    Nope

  7. RajshikharGupta
    • 3 years ago
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    expand the root term by binomial theorem

  8. RajshikharGupta
    • 3 years ago
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    then there will be term like 10x/sinx and x/sinx=1 for given condition so ans =10

  9. RajshikharGupta
    • 3 years ago
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    while opening with binomial neglect higher degree terms

  10. Valdas
    • 3 years ago
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    well, the answer should be 0.4 . And what do you mean by expanding the root term by binomial theorem? Do you mean, that I should raise the fraction by the 5th degree and then use the binomial theorem and numerator?

  11. RajshikharGupta
    • 3 years ago
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    no see it i have shown|dw:1353844296253:dw|

  12. RajshikharGupta
    • 3 years ago
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    and i have done mistake while doing this it is 2/5=0.4 so sorry for that

  13. Valdas
    • 3 years ago
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    Could you explain how did you get \[\sqrt[5]{1+2n}=1+\frac{ 2 }{ 5 }n + ....?\]

  14. RajshikharGupta
    • 3 years ago
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    it is binomial theorem u will learn this in algebra in high school it is (1+x)^n=1+(nC1)x+(nC2)x^2+(nC3)x^3 and so on .....

  15. Valdas
    • 3 years ago
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    Yes, I know the binomial theorem, but how does it apply to roots? I thought that the power of polynomial it is raised to must be an integer to apply this theorem.

  16. RajshikharGupta
    • 3 years ago
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    no it can be applied on roots if x is very small and it can even for any fractional powers for same case it quite valid approximation tool in maths and physics u can conform it with ur teachers

  17. Valdas
    • 3 years ago
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    Perhaps anyone can think of other way to find the limit? Every problem until this one required some sort of quite simple algebraic manipulation.

  18. RajshikharGupta
    • 3 years ago
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    actually it is one of the shortest methods but indeed u can initiate in ur problem by factorizing the numerator term

  19. pasta
    • 3 years ago
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    Binomial theorem just makes work hard ,learn l hopitals rule it is easy.LOOK IF after substituting the limit you get 0/0 differentiate the numenartor and denominator independently the find the limit three steps

  20. Valdas
    • 3 years ago
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    I can't use l'hoptial's rule to find the limit

  21. pasta
    • 3 years ago
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    i do not get you.Do you mean ,you are not allowed to use l hopitals rule or you can not evaluate.

  22. Valdas
    • 3 years ago
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    We haven't studied about that rule yet

  23. pasta
    • 3 years ago
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    do u want to learn it?

  24. Valdas
    • 3 years ago
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    This problem can be solved without use of l'hopitals rule, I want to find out how. Perhaps there is some simple algebraic trick which could be used to find the limit or something like that

  25. shubhamsrg
    • 3 years ago
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    @RajshikharGupta 's method seems to be the shortest to me,

  26. pasta
    • 3 years ago
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    ok ,I SEE.I have two solutions so far ,binomial and l'hopitals.But looks like there are one's and may be try following the trigonometry side or otherwise am still checking

  27. shubhamsrg
    • 3 years ago
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    are you asking how do we get (1+x)^n = 1+xn (given |x|<1 ) ?

  28. Valdas
    • 3 years ago
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    Yeah, I don't underst that part

  29. shubhamsrg
    • 3 years ago
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    hmm,,heard of taylor series expansion ?

  30. Valdas
    • 3 years ago
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    No

  31. shubhamsrg
    • 3 years ago
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    then you might just wanna mug up this formulla : |dw:1353848320044:dw| you'll get to know how we get this when you learn taylor series..

  32. shubhamsrg
    • 3 years ago
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    make that substitution in your question.. you'll reach the ans directly then..

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