Question

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

Answer 1

In this type of question students have to apply the basic concepts of logarithm and limits. Here from the rules of natural log we have to use $\ln {{e}^{x}}=x$ and another one that is $\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)$ . Also we have to consider that the logarithm function is an increasing function. After that we have to use substitution for $\dfrac{{{e}^{x}}}{x}$ which gives us indeterminate form $\dfrac{\infty }{\infty }$ and hence we have to use L-Hospital’s rule as ${{e}^{x}}$ and x are differentiable everywhere.

Complete step by step solution:
Consider, $\displaystyle \lim_{x \to \infty }\left( x-\ln x \right)$
We know that, $\ln {{e}^{x}}=x$ . Hence, we can write,
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x-\ln x \right)=\displaystyle \lim_{x \to \infty }\left( \ln {{e}^{x}}-\ln x \right)$.
By applying $\ln a-\ln b=\ln \left( \dfrac{a}{b} \right)$ we get,
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x-\ln x \right)=\displaystyle \lim_{x \to \infty }\ln \left( \dfrac{{{e}^{x}}}{x} \right)$.
Let, $u=\dfrac{{{e}^{x}}}{x}$ and now take limit of u as x approaches to $\infty$:
$\Rightarrow \displaystyle \lim_{x \to \infty }u=\displaystyle \lim_{x \to \infty }\left( \dfrac{{{e}^{x}}}{x} \right)$.
As we get indeterminate form $\dfrac{\infty }{\infty }$ , also ${{e}^{x}}$ and x are differentiable everywhere, we can use L-Hospital’s Rule,
$\Rightarrow \displaystyle \lim_{x \to \infty }u=\displaystyle \lim_{x \to \infty }\left( \dfrac{{{e}^{x}}}{x} \right)=\displaystyle \lim_{x \to \infty }\dfrac{\dfrac{d}{dx}\left( {{e}^{x}} \right)}{\dfrac{d}{dx}x}=\displaystyle \lim_{x \to \infty }{{e}^{x}}=\infty$.
So we can say that, as $x \to \infty ,u\to \infty$ and hence,
$\Rightarrow \displaystyle \lim_{u\to \infty }\left( \ln u \right)=\infty$.
$\Rightarrow \displaystyle \lim_{u\to \infty }\left( \ln u \right)=\displaystyle \lim_{x \to \infty }\ln \left( \dfrac{{{e}^{x}}}{x} \right)=\displaystyle \lim_{x \to \infty }\left( \ln {{e}^{x}}-\ln x \right)=\displaystyle \lim_{x \to \infty }\left( x-\ln x \right)=\infty$.
$\Rightarrow \displaystyle \lim_{x \to \infty }\left( x-\ln x \right)=\infty$.
That means the limit of $\left( x-\ln x \right)$ as x approaches $\infty$ goes unbounded to $+\infty$.

Note: In this type of question students may make mistakes in converting x into a logarithm function by using $\ln {{e}^{x}}=x$ . Also students have to take care when they substitute the limits. If they directly substitute the limits of x at the step $\displaystyle \lim_{x \to \infty }\left( \ln {{e}^{x}}-\ln x \right)$ one may write it as $\left( \infty -\infty \right)=0$ which is wrong. The students have to take care in applying L-Hospital’s rule also. By L-Hospital’s rule, if we get an indeterminate form then we have to take the derivative of the numerator and denominator separately and don't use the division rule of derivative.