Question

1. (4 points) find the following limit: \\( \lim _{n \

ightarrow \infty} \frac{\sinh (3 x)}{\cosh (2 x)} \\). fully justify your answer.

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 \(e^{2x}\)

$$
LATEXBLOCK0
$$

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:

\(\infty\)