## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{x\to0^+}\sin\Big(\frac{\pi}{2}e^{\sqrt x}\Big)=1$$ The function is continuous at $x=0$ from the right.
*Recall Theorem 10: If $g$ is continuous at $b$ and $\lim_{x\to c}f(x)=b$, then $$\lim_{x\to c}g(f(x))=g(b)=g(\lim_{x\to c}f(x))$$ $$A=\lim_{x\to0^+}f(x)=\lim_{x\to0^+}\sin\Big(\frac{\pi}{2}e^{\sqrt x}\Big)$$ - With functions $y=\sqrt x$: For all $x\ge0$, $\lim_{x\to c}\sqrt x=\sqrt c$, so $y=\sqrt x$ is continuous on $[0,\infty)$ - With functions $y=\sin x$ and $y=e^x$: For all $x\in R$, $\lim_{x\to c}\sin x=\sin c$ and $\lim_{x\to c}e^x=e^c$, so both functions are continuous on $(-\infty,\infty)$ That means we can apply Theorem 10 here as long as $\lim_{x\to0^+}x\ge0$ (so that $\sqrt x$ is defined). In detail, $$A=\sin\Big(\frac{\pi}{2}e^{\sqrt{\lim_{x\to0^+}x}}\Big)$$ $$A=\sin\Big(\frac{\pi}{2}e^{\sqrt{0}}\Big)=f(0)$$ $$A=\sin\Big(\frac{\pi}{2}\times1\Big)$$ $$A=\sin\frac{\pi}{2}=1$$ As shown above, $\lim_{x\to0^+}f(x)=f(0)$, so the function is continuous at $x=0$ from the right.