Question of limit 1/log(x without lhopital $\lim_{x\to\infty}{{\log \log \left(1+{{1}\over{x}}\right)}\over{\log x}}$
I cant do it. If somebody can do the question for me. I need do to the test but I not allowed use l'hopital.
 A: Since
$\log(1+1/x)
\approx 1/x
$
for large $x$
(easy to show from
$\log(1+z)
=\int_1^{1+z} \frac{dy}{y}
$,
from which you get
$\frac{z}{1+z} < \log(1+z) < z$),
${{\log \log \left(1+{{1}\over{x}}\right)}\over{\log x}}
\approx {{\log (1/x)}\over{\log x}}
= {-\log (x)\over{\log x}}
\to -1
$.
A: By multiplying x,
$\displaystyle\lim \limits_{x \to \infty} \frac{\log\log(1+\frac{1}{x})}{\log x} = \lim \limits_{x \to \infty} \frac{\log \frac{x\log (1+1/x)}{x}}{\log x} = \lim \limits_{x \to \infty} \frac{\log\log(1+1/x)^x-\log x}{\log x} = (\log\log e) (0) - 1$
$=-1$
A: 

In THIS ANSWER, I used on the limit definition of the exponential function and Bernoulli's Inequality to show that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\le \log(x)\le x-1 \tag 1$$
for $x>0$.


Using $(1)$, it is straightforward to show that
$$\frac{-x}{x-1}\le \frac{\log\left(\log\left(1+\frac1x\right)\right)}{\log(x)}\le \frac{1/x-1}{\frac{x-1}{x}}=-1$$
whereupon application of the squeeze theorem yields the coveted limit
$$\lim_{x\to \infty} \frac{\log\left(\log\left(1+\frac1x\right)\right)}{\log(x)}=-1$$

Note that we did not rely on L'Hospital's Rule, expansions, or other tools from differential calculus.  Rather, we only used the inequality in $(1)$, which was obtained from elementary analysis, and the squeeze theorem.

A: Use equivalents:
$\log(1+u)\sim_0 u$, hence $\log\Bigl(1+\dfrac 1x\Bigr)\sim_\infty\dfrac1x$. As it doesn't approach $1$, there results that
$$\log\log\Bigl(1+\dfrac 1x\Bigr)\sim_\infty -\log x,\enspace\text{whence}\quad \frac{\log\log\Bigl(1+\dfrac 1x\Bigr)}{\log x}\sim_\infty\frac{-\log x}{\log x}=-1.$$
