use the squeeze theorem to evaluate the limit: $lim_{x\rightarrow0 +}sqrt{7x}cdot e^{cos(3pi/x)}=$

Explanation:

Step1: Determine the range of cosine function

We know that $- 1\leqslant\cos\left(\frac{3\pi}{x}\right)\leqslant1$ for $x\neq0$. Then, since the exponential - function $y = e^{u}$ is an increasing function, we have $e^{-1}\leqslant e^{\cos\left(\frac{3\pi}{x}\right)}\leqslant e^{1}$ for $x\neq0$.

Step2: Multiply by $\sqrt{7x}$

Multiply each part of the inequality by $\sqrt{7x}$ (where $x>0$ as $x\rightarrow0^{+}$). We get $\sqrt{7x}e^{-1}\leqslant\sqrt{7x}e^{\cos\left(\frac{3\pi}{x}\right)}\leqslant\sqrt{7x}e^{1}$.

Step3: Find the limits of the bounding functions

Calculate $\lim_{x\rightarrow0^{+}}\sqrt{7x}e^{-1}=e^{-1}\lim_{x\rightarrow0^{+}}\sqrt{7x}=0$ and $\lim_{x\rightarrow0^{+}}\sqrt{7x}e^{1}=e^{1}\lim_{x\rightarrow0^{+}}\sqrt{7x}=0$.

Step4: Apply the Squeeze Theorem

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 $\lim_{x\rightarrow a}f(x)=\lim_{x\rightarrow a}h(x)=L$, then $\lim_{x\rightarrow a}g(x)=L$. Here, $a = 0^{+}$, $f(x)=\sqrt{7x}e^{-1}$, $g(x)=\sqrt{7x}e^{\cos\left(\frac{3\pi}{x}\right)}$, and $h(x)=\sqrt{7x}e^{1}$. So, $\lim_{x\rightarrow0^{+}}\sqrt{7x}e^{\cos\left(\frac{3\pi}{x}\right)} = 0$.

Answer:

$0$