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Home/ Questions/Q 3193
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nalakack
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nalakackBegginer ( Level 6 )
Asked: April 29, 20232023-04-29T21:08:46+05:30 2023-04-29T21:08:46+05:30In: Calculus ( කලනය )

Limits at infinity

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What is the limiting value of the xth root of x when x tends to infinity?

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  1. Gaya Nayomi
    Gaya Nayomi Level 1 ( Top Rated ) Senior Lecturer-University of Sri Jayewardenepura
    2023-05-05T22:08:05+05:30Added an answer on May 5, 2023 at 10:08 pm

    What do other teachers think about this question? Let us have a conversation around this question first. We will then give you an answer.

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  2. Janaka Rodrigo
    Janaka Rodrigo Level 2 English medium advanced level mathematics teacher at St Benedicts College Colombo 13
    2023-05-21T19:38:10+05:30Added an answer on May 21, 2023 at 7:38 pm
    This answer was edited.

    Since L’hopital’s theorem is not relevant to our local advanced level syllabus, I would like to suggest following approach,

    x^(1/x) = e^(ln(x^(1/x))) = e^( (lnx)/x)  ___ (1)

    Therefore you have to find limit of (lnx)/x

    0 < lnx < x and x – lnx is increasing function as x increases without bound when x > 1 . ( by differentiating x – lnx you can easily prove that)

    Therefore 0 < (lnx)/x <1 and since x – lnx is increasing function (lnx )/x must be decreasing function as x increases and that implies limit of (lnx)/x as x tends to infinity should be 0 .

    Now you can substitute the result to equation (1) and it gives the limit e^0 that is 1.

    As an alternative approach you can use the substitution lnx = t and prove that the limit of (lnx)/x as x tends to infinity is equal to the limit of t/( e^t) as t tends to infinity and the result is same.

     

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  3. Gaya Nayomi
    Gaya Nayomi Level 1 ( Top Rated ) Senior Lecturer-University of Sri Jayewardenepura
    2023-08-17T13:26:01+05:30Added an answer on August 17, 2023 at 1:26 pm

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  4. Gaya Nayomi
    Gaya Nayomi Level 1 ( Top Rated ) Senior Lecturer-University of Sri Jayewardenepura
    2023-08-17T13:26:42+05:30Added an answer on August 17, 2023 at 1:26 pm

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