# Question 7a02c

Dec 27, 2017

${\lim}_{x \to \infty} {\left(\frac{\log | x |}{x}\right)}^{\frac{1}{x}} = 1$

#### 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)

${\lim}_{x \to \infty} x = \infty$

${\lim}_{x \to \infty} \log \left(\left\mid x \right\mid\right) = \infty$

This gives us:

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

$\frac{\cancel{{\lim}_{x \to \infty} {\left(\infty\right)}^{\frac{1}{x}}}}{\cancel{{\lim}_{x \to \infty} {\left(\infty\right)}^{\frac{1}{x}}}} = 1$

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 $f \left(x\right) = {\left(\frac{\ln | x |}{x}\right)}^{\frac{1}{x}}$ (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 $\ln \left(f \left(x\right)\right) = \frac{1}{x} \cdot \ln \left(\frac{\ln | x |}{x}\right) = \frac{\ln \left(\ln | x |\right) - \ln \left(x\right)}{x}$, by properties of logarithms.

The limit ${\lim}_{x \to \infty} \ln \left(f \left(x\right)\right)$ is thus an $\frac{\infty}{\infty}$ indeterminate form , to which L'Hopital's Rule can be applied.

Since $\frac{d}{\mathrm{dx}} \left(\ln \left(\ln | x |\right) - \ln \left(x\right)\right) = \frac{1}{\ln | x |} \cdot \frac{1}{x} - \frac{1}{x}$ and $\frac{d}{\mathrm{dx}} \left(x\right) = 1$, we can say that

${\lim}_{x \to \infty} \ln \left(f \left(x\right)\right) = {\lim}_{x \to \infty} \frac{\frac{1}{\ln | x |} \cdot \frac{1}{x} - \frac{1}{x}}{1} = \frac{0}{1} = 0$.

The continuity of the natural log function can now be used to say that this means $\ln \left({\lim}_{x \to \infty} f \left(x\right)\right) = 0$ so that ${\lim}_{x \to \infty} f \left(x\right) = {e}^{0} = 1$.