\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{lnn}}}}{{\sqrt {\rm{n}} }}\)
Use \({\rm{a ln b = ln }}{{\rm{b}}^{\rm{a}}}\),
\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{ln}}{{\left( {{{\rm{n}}^{{\rm{1/2}}}}} \right)}^{\rm{2}}}}}{{\sqrt {\rm{n}} }}{\rm{n}}\\\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } \frac{{{\rm{2ln}}\sqrt {\rm{n}} }}{{\sqrt {\rm{n}} }}\end{array}\)
Substitute \(\sqrt {\rm{n}} {\rm{ = t}}\)if \({\rm{t}} \to \infty \)when \({\rm{n}} \to \infty \).
\(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{t}} \to \infty } \frac{{{\rm{2lnt}}}}{{{\rm{2t}}}}\)
So, the limit is \(\frac{\infty }{\infty }\).
L'Hospital's Rule
L'Hospital's Rule tells us that if can an indeterminate form\({\rm{0/0}}\)or\(\infty {\rm{/}}\infty \)all need to do is differentiate the numerator and differentiate the denominator and then take the limit
Implies this rule,
\(\begin{array}{c}\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{t}} \to \infty } \frac{{\rm{2}}}{{\rm{t}}}\\{\rm{ = }}\mathop {{\rm{lim}}}\limits_{{\rm{t}} \to \infty } \frac{{\rm{2}}}{{\rm{t}}}\\\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = 0}}\end{array}\)
Therefore, the given sequence converges to \(\mathop {{\rm{lim}}}\limits_{{\rm{n}} \to \infty } {{\rm{a}}_{\rm{n}}}{\rm{ = 0}}\).
