## Calculus: Early Transcendentals 8th Edition

Apply the squeeze theorem to prove that $\lim\limits_{x\to0^+}\sqrt xe^{\sin(\frac{\pi}{x})}=0$.
1) We know that $$-1\leq \sin(\frac{\pi}{x})\leq1$$ 2) For $\geq1$, function $e^x$ increases as $x$ increases. So, $$e^{-1}\leq e^{\sin(\frac{\pi}{x})}\leq e^1$$ 3) Multiply by $\sqrt x$ throughout. The inequality direction would not change, since $\sqrt x\geq0$ for $\forall x\in R$ $$\sqrt xe^{-1}\leq\sqrt xe^{\sin(\frac{\pi}{x})}\leq\sqrt xe^1$$ 4) Now consider the following limits $\lim\limits_{x\to0^+}\sqrt xe^{-1}=\sqrt0\times e^{-1}=0$ $\lim\limits_{x\to0^+}\sqrt xe^1=\sqrt0\times e^1=0$ Therefore, $\lim\limits_{x\to0^+}\sqrt xe^{-1}=\lim\limits_{x\to0^+}\sqrt xe^1=0$ 5) Applying the sqeeze theorem to the findings in 3) and 4), we find that $$\lim\limits_{x\to0^+}\sqrt xe^{\sin(\frac{\pi}{x})}=0$$