Question of limit 1/log(x without lhopital

Question

$\lim_{x\to\infty}{{\log \log \left(1+{{1}\over{x}}\right)}\over{\log x}}$

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.

Use the fact that for any $\alpha\in (0,1)$ there exists $M_{\alpha}>0$ such that, for all $x>M_{\alpha}$, $$\frac\alpha x\le \log\left(1+\frac1x\right)\le \frac1x$$ — , May 19 '16 at 23:54
I am referring to the statement of the problem. From what has been stated, $\alpha$ could be an arbitrary real value, i.e. $\alpha \in \mathbb{R}.$ @G.Sassatelli — Sergio Charles, May 20 '16 at 00:15
@multivariablecalculus $\alpha$ is a number arbitrarily close to $1$. Think of $1-\varepsilon$. — Heimdall, May 20 '16 at 00:16
@Multivariablecalculus It was a questionable choice of latex ( {\it \propto} instead of \infty ). — , May 20 '16 at 00:19
If we put $1/x = t$ we can see that $$\dfrac{\log\log(1 + (1/x))}{\log x} = -\frac{\log\log(1 + t)}{\log t} = -\frac{\log((1/t)\log(1 + t)) + \log t}{\log t} = -1 - \frac{\log((1/t)\log (1 + t))}{\log t} \to -1 + 0$$ as $ t \to 0$. — Paramanand Singh, May 21 '16 at 02:24

score 1 · Answer 1 · answered May 20 '16 at 00:14

1

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 $.

answered May 20 '16 at 00:14

marty cohen

107,799

you use lhopital? i dont understand Integral because i havent this matery yet – user340886 May 20 '16 at 00:21
The integral is how log is defined. L'hopital is not used at all. – marty cohen May 20 '16 at 07:34

score 1 · Answer 2 · answered May 20 '16 at 01:41

By multiplying x,

$\displaystyle\lim \limits_{x \to \infty} \frac{\log\log(1+\frac{1}{x})}{\log x} = \lim \limits_{x \to \infty} \frac{\log \frac{x\log (1+1/x)}{x}}{\log x} = \lim \limits_{x \to \infty} \frac{\log\log(1+1/x)^x-\log x}{\log x} = (\log\log e) (0) - 1$

$=-1$

score 1 · Answer 3 · edited Jun 12 '20 at 10:38

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.

score 0 · Answer 4 · answered May 20 '16 at 01:16

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 $$\log\log\Bigl(1+\dfrac 1x\Bigr)\sim_\infty -\log x,\enspace\text{whence}\quad \frac{\log\log\Bigl(1+\dfrac 1x\Bigr)}{\log x}\sim_\infty\frac{-\log x}{\log x}=-1.$$

Question of limit 1/log(x without lhopital

4 Answers4