6. Use the Squeeze Theorem to prove the following: (a) \( \lim _{x \rightarrow 0} \sqrt{x^{3}+x^{2}} \cos \left(\frac{\pi}{x}\right)=0 \) (b) \( \lim _{x \rightarrow 0^{+}} \sqrt{x} e^{\sin \left(\frac{\pi}{x}\right)}=0 \)

**(a)** We wish to evaluate \[ \lim_{x \rightarrow 0} \sqrt{x^3 + x^2} \cos\left(\frac{\pi}{x}\right). \] 1. Notice that for all real \( x \) (and particularly for \( x \neq 0 \)), we have \[ -1 \leq \cos\left(\frac{\pi}{x}\right) \leq 1. \] 2. Multiplying the inequality by the nonnegative quantity \( \sqrt{x^3+x^2} \) gives \[ -\sqrt{x^3+x^2} \leq \sqrt{x^3+x^2}\cos\left(\frac{\pi}{x}\right) \leq \sqrt{x^3+x^2}. \] 3. Consider the limit of the bounding functions as \( x \rightarrow 0 \): \[ \lim_{x \rightarrow 0} \sqrt{x^3+x^2} = 0. \] This is because we can factor the expression as \[ \sqrt{x^3+x^2} = \sqrt{x^2(x+1)} = |x|\sqrt{x+1}, \] and as \( x \rightarrow 0 \), both \( |x| \) and \( \sqrt{x+1} \) tend to \( 0 \) and \( 1 \), respectively. 4. By the Squeeze Theorem, since both the lower bound \( -\sqrt{x^3+x^2} \) and the upper bound \( \sqrt{x^3+x^2} \) tend to \( 0 \) as \( x \rightarrow 0 \), it follows that \[ \lim_{x \rightarrow 0} \sqrt{x^3+x^2}\cos\left(\frac{\pi}{x}\right)=0. \] --- **(b)** We wish to evaluate \[ \lim_{x \rightarrow 0^{+}} \sqrt{x}\, e^{\sin\left(\frac{\pi}{x}\right)}. \] 1. Again, note that for all \( x \), \[ -1 \leq \sin\left(\frac{\pi}{x}\right) \leq 1. \] 2. Exponentiating the inequality (using the fact that the exponential function is increasing) yields \[ e^{-1} \leq e^{\sin\left(\frac{\pi}{x}\right)} \leq e^1. \] 3. Multiplying through by \( \sqrt{x} \) (which is nonnegative for \( x>0 \)) gives \[ e^{-1}\sqrt{x} \leq \sqrt{x}\, e^{\sin\left(\frac{\pi}{x}\right)} \leq e\sqrt{x}. \] 4. Compute the limits of the bounding functions as \( x \rightarrow 0^+ \): \[ \lim_{x \rightarrow 0^+} e^{-1}\sqrt{x} =0 \quad \text{and} \quad \lim_{x \rightarrow 0^+} e\sqrt{x} =0. \] 5. By the Squeeze Theorem, \[ \lim_{x \rightarrow 0^{+}} \sqrt{x}\, e^{\sin\left(\frac{\pi}{x}\right)}=0. \]