Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\frac{\ln (x)}{x}$ using sandwich Theorem (Squeeze theorem).

$\bf{My\ Try:}$ Let $\ln (x) = y\Rightarrow x=e^y$ and when $x\rightarrow \infty, y=\ln(x)\rightarrow \infty$

$\displaystyle \lim_{y\rightarrow \infty}\frac{y}{e^y}$, Now $\displaystyle e^y = 1+\frac{y}{1!}+\frac{y^2}{2!}+\frac{y^3}{3!}+..........+\infty$

Now I did not understand how I calculate The Given limit using Squeeze theorem

Help required



For any positive, real $k$ there is a $c$ such that $kx + c \gt \ln(x)$ (a proof of this is below using derivatives). If you divide each side by $x$, you can conclude that your function is less than $k + c/x$.

So we have $$ k + \frac{c}{x} \gt\frac{ \ln(x)}{x} \gt 0 $$ Remember that this is really a load of squeezing inequalities, one for each choice of $k$, with a fitting $c$ to go with it.

In the limit as $x \to \infty$, this becomes $$ k \geq \lim_{x \to \infty}\frac{\ln(x)}{x} \geq 0 $$ Since $k$ could be any positive, real number, we have the result we want.

Proof of the existence of $c$:

Let's set $x_0 = \frac{1}{k}$. I claim that any $c$ greater than $$ c' = \ln(x_0) - kx_0 $$ works. We see that $kx + c'$ is tangent to $\ln(x)$ at $x = x_0$, since they have the same functional value and the same derivative. They also do not intersect at any other point since at any point before $x_0$, $\ln(x)$ has greater derivative, and at any point after $kx + c'$ has the greater derivative.

Therefore, any $c\gt c'$ will result in a line $kx + c$ which is strictly greater than $\ln(x)$ for all positive $x$.

  • $\begingroup$ Can the non-negativeness of $f(x)=x^{r}-r\ln x$ for $x\geq 1$, where $0<r<1$, be sufficient? If it does, then we have $0\leq\frac{\ln x}{x}\leq \frac{1}{rx^{1-r}}$ and hence the Sq.theorem is applied. $\endgroup$ – daulomb Oct 29 '13 at 11:19
  • $\begingroup$ @user40615 That is more fitting as a comment to the question, not to my answer. But yeah, that should work too. You just need to take care to specify where your $f_r$ is non-negative, since it isn't all of $x \geq 1$ for small enough $r$. You need to prove that it eventually gets positive, or correct it with a constant term as I did with my $c$. $\endgroup$ – Arthur Oct 29 '13 at 11:35
  • $\begingroup$ Sorry Arthur it was not a comment to your answer. $f$ must have been $f(x)=x^r-r\ln x-1$. My comment actually contais a question: since $x$ goes to infinity, could we use any value of $x$ which makes $f$ positive and apply the squeezing. $\endgroup$ – daulomb Oct 29 '13 at 14:38

Let us put $x = e^{y} $ as OP has done then we see that $$\lim_{x \to \infty}\frac{\ln x}{x} = \lim_{y \to \infty}\frac{y}{e^{y}}$$ and since OP is familiar with series expansion of $e^{y}$ things are much simpler now. We have $$e^{y} = 1 + y + \frac{y^{2}}{2!} + \cdots > \frac{y^{2}}{2!}$$ We now have $y/e^{y} < 2/y$ and thus we see that $$0 < \frac{y}{e^{y}} < \frac{2}{y}$$ Using squeeze theorem we see that $$\lim_{y \to \infty} 0 \leq \lim_{y \to \infty}\frac{y}{e^{y}} \leq \lim_{y \to \infty}\frac{2}{y} = 0$$ and hence $$\lim_{y \to \infty}\frac{y}{e^{y}} = 0$$ It is also possible to use a direct inequality regarding $\ln x$. Let $a \in (0, 1)$ then we know that we have $t^{-1} < t^{a - 1}$ for all $t \in (1, \infty)$ and hence we must have $$\int_{1}^{x}\frac{dt}{t} < \int_{1}^{x}t^{a - 1}\,dt$$ i.e. $$\ln x < \frac{x^{a} - 1}{a}$$ for all $x > 1$. So we get $$0 < \frac{\ln x}{x} < \frac{x^{a} - 1}{ax}$$ Using squeeze theorem we get $$\lim_{x \to \infty}0 \leq \lim_{x \to \infty}\frac{\ln x}{x} \leq \lim_{x \to \infty}\frac{x^{a} - 1}{ax} = 0$$ The last limit is $0$ because $a \in (0, 1)$ (or one can put $a = 1/2$ and more easily see that last limit is $0$).


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