Question

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

Answer 1

Hint: 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.