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
Thanks
Best Answer
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 $x$.