# How to solve this limit ?

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

Note by Brilliant Member
1 year, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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

- 1 year, 6 months ago