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