Question

Prove that \(\mathop {{\bf{lim}}}\limits_{x \to {{\bf{0}}^ + }} \sqrt x {e^{{\bf{sin}}\left( {\frac{\pi }{x}} \right)}} = 0\).

Short answer

Hence, it is proved that \(\mathop {lim}\limits_{x \to {0^ + }} \sqrt x {e^{sin\left( {\frac{\pi }{x}} \right)}} = 0\).

Step-by-step solution

Find the value of the expression using the range of \({\bf{sin}}\frac{\pi }{x}\)

The range of a function \(\sin \frac{\pi }{x}\) is:

\(\begin{array}{c} - 1 \le \sin \frac{\pi }{x} \le 1\\{e^{ - 1}} \le {e^{\sin \frac{\pi }{x}}} \le {e^1}\\\frac{{\sqrt x }}{e} \le \sqrt x {e^{\sin \frac{\pi }{x}}} \le \sqrt x e\end{array}\)

Evaluate the limit using the squeeze theorem

Solve the expression for limit using the squeeze theorem.

\(\mathop {\lim }\limits_{x \to {0^ + }} \left( {\frac{{\sqrt x }}{e}} \right) = 0\)and \(\mathop {\lim }\limits_{x \to {0^ + }} \left( {\sqrt x e} \right) = 0\)

Since \(\mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sqrt x }}{e}} \right) = 0\), by the squeeze theorem we have:

\(\mathop {{\bf{lim}}}\limits_{x \to {\bf{0}}} \sqrt x {e^{sin\left( {\frac{\pi }{x}} \right)}} = 0\)

Thus, the value of the limit is \(\mathop {{\bf{lim}}}\limits_{x \to {\bf{0}}} \sqrt x {e^{sin\left( {\frac{\pi }{x}} \right)}} = 0\).