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

34. \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - x}} + 2{\rm{cos}}3x} \right)\)

Short Answer

Expert verified

The limit \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - x}} + 2{\rm{cos}}3x} \right)\) does not exist.

Step by step solution

01

Determine the limit of each function

Consider the limit \(\mathop {\lim }\limits_{x \to \infty } \left( {{e^{ - x}} + 2\cos 3x} \right)\).

The limit of \({e^{ - x}}\) at \(x\) tends to \(\infty \) is 0, that is, \(\mathop {{\rm{lim}}}\limits_{x \to \infty } {e^{ - x}} = 0\).

The limit of \(2\cos 3x\) does not exist because, as \(x\) tends to \(\infty \), the value of \(2\cos 3x\) oscillating infinitely between the values of \( - 2\) and \(2\).

Thus, \(\mathop {\lim }\limits_{x \to \infty } \left( {2\cos 3x} \right)\) does not exist.

02

Find the limit of the given function

As \(\mathop {{\rm{lim}}}\limits_{x \to \infty } 2{\rm{cos}}3x\) does not exist, this implies that \(\mathop {{\rm{lim}}}\limits_{x \to \infty } \left( {{e^{ - x}} + 2{\rm{cos}}3x} \right)\) does not exist.

Therefore, the limit of the given function does not exist.

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