5. Show that \( \lim _{x \rightarrow 9} \sqrt{x}=3 \). 6. Evaluate the following limits using the Squeeze Theorem a) \( \lim _{x \rightarrow 0} x^{4} \sin \left(\frac{7}{x}\right)+5 \).

To show that \( \lim _{x \rightarrow 9} \sqrt{x}=3 \), we can use the direct substitution method. As \( x \) approaches 9, substituting \( x = 9 \) into the square root gives \( \sqrt{9} = 3 \). Since this is a continuous function, the limit follows directly as \( x \) approaches 9. For the limit \( \lim _{x \rightarrow 0} x^{4} \sin \left(\frac{7}{x}\right) + 5 \), we can apply the Squeeze Theorem. We know that \( -1 \leq \sin \left(\frac{7}{x}\right) \leq 1 \), hence \( -x^4 \leq x^4 \sin \left(\frac{7}{x}\right) \leq x^4 \). As \( x \) approaches 0, both \( -x^4 \) and \( x^4 \) approach 0. Therefore, by the Squeeze Theorem, \( \lim _{x \rightarrow 0} x^{4} \sin \left(\frac{7}{x}\right) = 0 \). Adding 5 gives \( 5 \). Thus, the overall limit is \( 5 \).