Excel in math, science, and engineering

New user? Sign up

Existing user? Sign in

\[ \large \lim_{n\to\infty} \ln | \sinh n | = \, ? \]

Note by Soma Arjun 6 months, 4 weeks ago

Sort by:

The limit goes to infinity. Consider:

\(\sinh { \left( n \right) } =\frac { { e }^{ n }-{ e }^{ -n } }{ 2 } \)

We can say that: \(\displaystyle \lim _{ n\rightarrow \infty }{ \left( { e }^{ -n } \right) } =0\)

Therefore, \(\displaystyle \lim _{ n\rightarrow \infty }{ \left( \sinh { \left( n \right) } \right) } =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ n } }{ 2 } \right) } =\lim _{ n\rightarrow \infty }{ \left( n \right) } -\ln { \left( 2 \right) } \) – Aditya Kumar · 6 months, 3 weeks ago

Log in to reply

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThe limit goes to infinity. Consider:

\(\sinh { \left( n \right) } =\frac { { e }^{ n }-{ e }^{ -n } }{ 2 } \)

We can say that: \(\displaystyle \lim _{ n\rightarrow \infty }{ \left( { e }^{ -n } \right) } =0\)

Therefore, \(\displaystyle \lim _{ n\rightarrow \infty }{ \left( \sinh { \left( n \right) } \right) } =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ n } }{ 2 } \right) } =\lim _{ n\rightarrow \infty }{ \left( n \right) } -\ln { \left( 2 \right) } \) – Aditya Kumar · 6 months, 3 weeks ago

Log in to reply