Question

To determine the value of \(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right)\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right)\) is .

Step-by-step solution

Step 2: Concept of limits

Limit, mathematical concept based on the idea of closeness, used primarily to assign values to certain functions at points where no values are defined, in such a way as to be consistent with nearby values.

Obtain the value of \(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right)\)

The value of\(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right)\)is shown below.

\(\begin{array}{c}\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right) = \left( {{e^{ - 2(\infty )}}\cos (\infty )} \right)\\ = {e^{ - \infty }} \times \cos (\infty )\end{array}\)

Since,\(\cos (\infty ) \in \left( { - 1,1} \right)\)and\({e^{ - \infty }} = 0\), so

\(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right) = 0\)

So, the value of \(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - 2x}}\cos x} \right)\) is 0.