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# How to solve this limit ?

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

Note by Soma Arjun
4 months, 4 weeks ago

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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, $$\displaystyle \lim _{ n\rightarrow \infty }{ \left( \sinh { \left( n \right) } \right) } =\lim _{ n\rightarrow \infty }{ \left( \frac { { e }^{ n } }{ 2 } \right) } =\lim _{ n\rightarrow \infty }{ \left( n \right) } -\ln { \left( 2 \right) }$$ · 4 months, 2 weeks ago