Question

15–42: Find the limit or show that it does not exist.

35.\(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - 2x}}{\rm{cos}}x} \right)\)

Short answer

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

Step-by-step solution

Find the limit

Since the value of \({\rm{cos}}x\) lies between \( - 1\) and \(1\), that is, \( - 1 \le {\rm{cos}}x \le 1\).

Multiply each side of the inequality by \({e^{ - 2x}}\) and simplify it.

\(\begin{array}{c} - 1\left( {{e^{ - 2x}}} \right) \le {e^{ - 2x}}{\rm{cos}}x \le 1\left( {{e^{ - 2x}}} \right)\\ - {e^{ - 2x}} \le {e^{ - 2x}}{\rm{cos}}x \le {e^{ - 2x}}\end{array}\)

Apply Squeeze Theorem

It is known that the limit of \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( { - {e^{ - 2x}}} \right) = 0\), and \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - 2x}}} \right) = 0\).

By using the Squeeze Theorem, the limit of \({e^{ - 2x}}{\rm{cos}}x\) as \(x\) tends to \(\infty \) is \(0\), that is, \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - 2x}}{\rm{cos}}x} \right) = 0\).

Thus, the value of the limit is 0.