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**Hint:**As we know that the above question is an example of limit function. A limit of a function can be defined as a number that a function reaches as the independent variable of the function reaches at a given number. The limit of a real valued function ”$f$” with respect to the variable $x$ can be defined as the $\mathop {\lim }\limits_{x \to p} f(x) = L$. Here lim refers to the limit of the function. We can say that the limit of any given function $'f'$ of $x$ as $x$ approaches to $p$ is equal to $L$. L’Hopital’s rule gives us the method to evaluate limits of indeterminate forms.

**Complete step by step solution:**

Here we have $x - 1n(x)$, we can write it as $1n({e^x}) + 1n({x^{ - 1}}) = 1n\left( {\dfrac{{{e^x}}}{x}} \right)$, here the exponential is infinity which is of great order. To be clear we will now take log at $z = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{e^x}}}{x}$ and by the rule the form which is indeterminate is $\dfrac{\infty }{\infty }$. Therefore by L’hopitals rule we have $\mathop {\lim }\limits_{x \to \infty } \dfrac{{{e^x}}}{x} = \mathop {\lim }\limits_{x \to \infty } \dfrac{{{e^1}}}{1} \to \infty $ and so by the above expression $z \to \infty $ and $1nz \to \infty $.

**Hence the required answer is that $1nz \to \infty $.**

**Note:**As we know that according to L’Hopital’s rule, the limit of a function when we divide one function by another function, it remains the same after we take the derivative of each function. The number $e$ is also known as Euler’s number which is a mathematical constant. It is also the base of the natural logarithm. It is defined by $e = \mathop {\lim }\limits_{n \to \infty } (1 + 1n)n$. When $x$ approaches infinity it means that there is no number that we can name.

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