Question

What is the limit as $x$ approaches infinity of $\ln \left( x \right)$?

Answer 1

For solving these types of questions you should know that the $\infty$ is not a value, so we take it as $x$ approaching infinity. And we should keep it always positive infinity because negative infinity’s logarithm is undefined for real numbers, since the natural logarithm function is undefined for negative numbers.

Complete step by step solution:
In this question it is given that $x$ approaches infinity. SO, it is clear that the limit must be positive because $x>\ln \left( x \right)>0$. By the properties of $\ln$, it is clear that,
$\ln \left( {{x}_{2}} \right)-\ln \left( {{x}_{1}} \right)=\ln \left( \dfrac{{{x}_{2}}}{{{x}_{1}}} \right)$
So, if ${{x}_{2}}>{{x}_{1}}$, the difference is positive, so the value of $\ln \left( x \right)$ is always growing. If we consider,
$\displaystyle \lim_{x \to \infty }\ln \left( x \right)=M$,
Then, $\displaystyle \lim_{x \to \infty }\ln \left( x \right)=M\in R$,
And we have, $\ln \left( x \right) < M \Rightarrow x < {{e}^{M}}$.
But the value of $x \to \infty$.
So, $M$ cannot be in the $R$ and the limits of this must be $+\infty$. Or if we see the mathematical calculation for this, then,
$\displaystyle \lim_{x \to \infty }\ln \left( x \right)=\infty$.
To see this, we will use the $\ln x=\int\limits_{1}^{x}{\dfrac{1}{t}dt}$ and by the property:
$\int\limits_{a}^{b}{f\left( t \right)dt}=\int\limits_{a}^{c}{f\left( t \right)dt}+\int\limits_{c}^{b}{f\left( t \right)dt}$
If on $\left[ a,b \right]$ we have $f\left( t \right)\ge m$
Then, $\int\limits_{a}^{b}{f\left( t \right)dt}\ge \left( b-a \right).m$
If we look at the intervals of $\left[ {{2}^{n}},{{2}^{n+1}} \right]$ form, then, on $\left[ 1,2 \right]$, we have $\dfrac{1}{t}\ge \dfrac{1}{2}$, so here,
$\begin{aligned} & \int\limits_{1}^{2}{\dfrac{1}{t}dt\ge \left( 2-1 \right).\dfrac{1}{2}} \\\ & \Rightarrow \int\limits_{1}^{2}{\dfrac{1}{t}dt}=\dfrac{1}{2} \\\ \end{aligned}$
So, $\ln 2\ge \dfrac{1}{2}$
And on $\left[ 2,4 \right]$, we have $\dfrac{1}{t}\ge \dfrac{1}{4}$, so,
$\begin{aligned} & \int\limits_{1}^{4}{\dfrac{1}{t}dt=\int\limits_{1}^{2}{\dfrac{1}{t}dt}+\int\limits_{2}^{4}{\dfrac{1}{t}dt}\ge \dfrac{1}{2}+\left( 4-2 \right).\dfrac{1}{4}} \\\ & \Rightarrow \int\limits_{1}^{4}{\dfrac{1}{t}dt}=\dfrac{1}{2}+\dfrac{1}{2}=1 \\\ \end{aligned}$
And, $\ln 4\ge \dfrac{2}{2}=1$
So, on every $\left[ {{2}^{n}},{{2}^{n+1}} \right]$, we have the values $\dfrac{1}{t}\ge \dfrac{1}{{{2}^{n}}+1}$.So, the integral $\int\limits_{{{2}^{n}}}^{{{2}^{n+1}}}{\dfrac{1}{t}dt}$ adds more than,
$\left( {{2}^{n+1}}-{{2}^{n}} \right).\dfrac{1}{{{2}^{n+1}}}=\left[ {{2}^{n}}\left( 2-1 \right) \right].\dfrac{1}{{{2}^{n+1}}}=\dfrac{{{2}^{n}}}{{{2}^{n+1}}}=\dfrac{1}{2}$
And so, $\ln \left( {{2}^{n+1}} \right)=\int\limits_{{{2}^{n}}}^{{{2}^{n+1}}}{\dfrac{1}{t}dt\ge \dfrac{n+1}{2}}$
So, if $x \to \infty$, we have $\int\limits_{1}^{x}{\dfrac{1}{t}dt\to \infty }$.
Since this integral is $\ln x$, we have $\displaystyle \lim_{x \to \infty }\ln \left( x \right)=\infty$.

Thus, the limit as $x$ approaches infinity of $\ln \left( x \right)$ is infinity.

Note: In this question we have to assume only positive infinite value, we cannot assume the negative infinite value. And on every ${{2}^{n,}}{{2}^{n+1}}$ values of a and b, the $\ln \left( x \right)$ function must be approaching to positive infinite. On every positive value of a and b, the $\ln \left( x \right)$ will be $x\to \infty$.