2. (4 points) find the following limit: \\( \\lim _{n \\rightarrow \\infty} \\frac{\\sinh (3 x)}{\\cosh (2…

Explanation:

Step1: Recall the definitions of hyperbolic functions

We know that (\sinh t=\frac{e^{t}-e^{-t}}{2}) and (\cosh t = \frac{e^{t}+e^{-t}}{2}). So, (\frac{\sinh(3x)}{\cosh(2x)}=\frac{\frac{e^{3x}-e^{-3x}}{2}}{\frac{e^{2x}+e^{-2x}}{2}}=\frac{e^{3x}-e^{-3x}}{e^{2x}+e^{-2x}}).

Step2: Divide numerator and denominator by (e^{2x})

[ \begin{align*} \frac{e^{3x}-e^{-3x}}{e^{2x}+e^{-2x}}&=\frac{e^{3x}\div e^{2x}-e^{-3x}\div e^{2x}}{e^{2x}\div e^{2x}+e^{-2x}\div e^{2x}}\ &=\frac{e^{x}-e^{-5x}}{1 + e^{-4x}} \end{align*} ]

Step3: Find the limit as (x\to\infty)

As (x\to\infty), (e^{-5x}=\frac{1}{e^{5x}}\to0) and (e^{-4x}=\frac{1}{e^{4x}}\to0). Then (\lim_{x\to\infty}\frac{e^{x}-e^{-5x}}{1 + e^{-4x}}=\lim_{x\to\infty}\frac{e^{x}-0}{1 + 0}). Since (\lim_{x\to\infty}e^{x}=\infty)

Answer:

(\infty)