Question

Question #7a02c

Answer 1

`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`

Answer 2

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`.