Question #7a02c

2 Answers
Dec 27, 2017

lim_(x to oo)((log|x|)/x)^(1/x)=1

Explanation:

This can be expressed as (lim_(x to oo)(log|x|)^(1/x))/(lim_(x to oo)(x)^(1/x))-=(lim_(x to oo)(lim_(x to oo)(log|x|))^(1/x))/(lim_(x to oo)(lim_(x to oo)x)^(1/x)

lim_(x to oo)x=oo

lim_(x to oo)log(abs(x))=oo

This gives us:

(lim_(x to oo)(oo)^(1/x))/(lim_(x to oo)(oo)^(1/x)

cancel(lim_(x to oo)(oo)^(1/x))/cancel(lim_(x to oo)(oo)^(1/x))=1

Dec 27, 2017

The answer is 1, but we should use a more rigorous approach, such as an application of L'Hopital's Rule .

Explanation:

Let f(x)=((ln|x|)/x)^(1/x) (I'm assuming you meant to use the natural logarithm here, though the approach used can be easily modified to use the common logarithm).

Then ln(f(x))=1/x * ln((ln|x|)/x)=(ln(ln|x|)-ln(x))/x, by properties of logarithms.

The limit lim_{x->infty}ln(f(x)) is thus an infty/infty indeterminate form , to which L'Hopital's Rule can be applied.

Since d/dx(ln(ln|x|)-ln(x))=1/(ln|x|)*1/x-1/x and d/dx(x)=1, we can say that

lim_{x->infty}ln(f(x))=lim_{x->infty}(1/(ln|x|)*1/x-1/x)/1=0/1=0.

The continuity of the natural log function can now be used to say that this means ln(lim_{x->infty}f(x))=0 so that lim_{x->infty}f(x)=e^{0}=1.