Question

Evaluate\(\mathop {lim}\limits_{x \to = } \frac{{sinhx}}{{{e^x}}}\).

Short answer

The value of \(\mathop {\lim }\limits_{x \to = } \frac{{\sinh x}}{{{e^x}}}\) is \(\frac{1}{2}\).

Step-by-step solution

Given limit

The given expression is\(\mathop {\lim }\limits_{x \to = } \frac{{\sinh x}}{{{e^x}}}\).

Concept of hyperbolic function

The hyperbolic function of\(sinhx\):\(sinhx = \frac{{{e^x} - {e^{ - x}}}}{2}\).

Evaluate the term and apply the limit

The given expression is \(\mathop {\lim }\limits_{x \to = } \frac{{\sinh x}}{{{e^x}}}\).

Evaluate the limit \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sinh x}}{{{e^x}}}\) as shown below.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \frac{{\sinh x}}{{{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {\frac{{{e^x} - {e^{ - x}}}}{2}} \right)}}{{{e^x}}}\quad \\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^x} - {e^{ - x}}}}{{2{e^x}}}} \right)\\ = \mathop {\lim }\limits_{x \to \infty } \left( {\frac{1}{2} - \frac{{{e^{ - 2x}}}}{2}} \right)\end{array}\)

Apply the limit to obtain the solution.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {\frac{{\sinh x}}{{{e^x}}}} \right) = \frac{1}{2} - \mathop {\lim }\limits_{x \to \infty } \left( {\frac{{{e^{ - 2x}}}}{2}} \right)\\ = \frac{1}{2} - \left( {\frac{{{e^{ - \infty }}}}{2}} \right)\\ = \frac{1}{2} - \frac{0}{2}\quad \\ = \frac{1}{2}\end{array}\)

Therefore, the value of \(\mathop {\lim }\limits_{x \to \infty } \frac{{\sinh x}}{{{e^x}}}\) is \(\frac{1}{2}\).