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

Sign Up to our MetaHub questions and Answers Engine to ask questions, answer members' questions, and connect with other teachers & members.

Login to our META-HUB questions & Answers Forum

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

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

~MetaHub Panel for Teachers’ Forum

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.