The answer must be not defined right?

Since direct substitution will get us $\dfrac{\infty}{\infty}$ , we can apply l'hopital's rule, which will get us $\displaystyle \lim_{x\rightarrow\infty}\dfrac{1}{x} = \boxed{0}$

L'hopital's Rule

Using L'hopital's Rule:

the given limit: lim of (lnx/x) as x ----> infinity

the given limit becomes lim of (1/x) as x ----> infinity = 0

as the x goes infinity the function becomes zero since 1 is constant doesn't get larger but x gets larger and larger thus it becomes 0

If u dont want to use L-Hospital, another method.... When x tends to infinity, both x and lnx tend to infinity.... But growth of lnx is very low (infact, it has d lowest growth rate amongst d basic functions), whereas growth rate of x is very high...... So we can directly say dat x is much larger dan lnx.... Hence, value of d limit is 0.....

L - Hopital Rule

just apply l hopital rule in numberator u will get 1/x and in denominator u will get 1 which equals to 1/x which is 0

simply 1 by infinity is zero