Question

assignment 3: problem 2 (1 point)
find $lim_{x
ightarrow0^{+}}sqrt{x}e^{sin(pi/x)}$

Explanation:

Step1: Analyze the range of sine function

We know that $- 1\leqslant\sin(\frac{\pi}{x})\leqslant1$ for $x
eq0$. Then $e^{-1}\leqslant e^{\sin(\frac{\pi}{x})}\leqslant e^{1}$.

Step2: Use the squeeze - theorem

We want to find $\lim_{x
ightarrow0^{+}}\sqrt{x}e^{\sin(\frac{\pi}{x})}$. Multiply the inequality by $\sqrt{x}$ (since $\sqrt{x}\geqslant0$ for $x\geqslant0$), we get $\sqrt{x}e^{-1}\leqslant\sqrt{x}e^{\sin(\frac{\pi}{x})}\leqslant\sqrt{x}e^{1}$.

Step3: Find the limits of the bounding functions

We calculate $\lim_{x
ightarrow0^{+}}\sqrt{x}e^{-1}=e^{-1}\lim_{x
ightarrow0^{+}}\sqrt{x} = 0$ and $\lim_{x
ightarrow0^{+}}\sqrt{x}e^{1}=e^{1}\lim_{x
ightarrow0^{+}}\sqrt{x}=0$.
By the squeeze - theorem, if $f(x)\leqslant g(x)\leqslant h(x)$ for all $x$ in some open interval containing $a$ (except possibly at $a$) and $\lim_{x
ightarrow a}f(x)=\lim_{x
ightarrow a}h(x) = L$, then $\lim_{x
ightarrow a}g(x)=L$. Here $a = 0^{+}$, $f(x)=\sqrt{x}e^{-1}$, $g(x)=\sqrt{x}e^{\sin(\frac{\pi}{x})}$, $h(x)=\sqrt{x}e^{1}$, and $L = 0$.

Answer:

$0$