10. Evaluate the following limit with full justification: \( \lim _{x \rightarrow \infty} \frac{2 x^{2}-6 x}{x^{3}+1} \) 11. Evaluate the following limit with full justification. [Hint: You will need the Squecze Theorem, which also works for limits at infinity.] \( \lim _{x \rightarrow \infty} \frac{\sin \left(x^{2}\right)}{x^{2}} \)

To evaluate the limit \( \lim _{x \rightarrow \infty} \frac{2 x^{2}-6 x}{x^{3}+1} \), we can first analyze the degrees of the polynomials in the numerator and denominator. The highest degree in the numerator is \(2\) (from \(2x^2\)) and in the denominator is \(3\) (from \(x^3\)). As \(x\) approaches infinity, the leading terms dominate the values of the polynomials. Thus, we simplify the limit as follows: \[ \lim _{x \rightarrow \infty} \frac{2 x^{2}-6 x}{x^{3}+1} = \lim _{x \rightarrow \infty} \frac{2}{x} - \frac{6}{x^2} \cdot \frac{1}{1+\frac{1}{x^3}} \] As \(x\) approaches infinity, both \(\frac{6}{x^2}\) and \(\frac{1}{1+\frac{1}{x^3}}\) approach \(0\). Therefore, the limit is: \[ 0. \] For the limit \( \lim _{x \rightarrow \infty} \frac{\sin \left(x^{2}\right)}{x^{2}} \), we will employ the Squeeze Theorem. The sine function is bounded, meaning \( -1 \leq \sin(x^2) \leq 1 \). Thus, we can write: \[ -\frac{1}{x^2} \leq \frac{\sin(x^2)}{x^2} \leq \frac{1}{x^2}. \] As \(x\) approaches infinity, both bounds \(-\frac{1}{x^2}\) and \(\frac{1}{x^2}\) converge to \(0\). By the Squeeze Theorem, we conclude: \[ \lim _{x \rightarrow \infty} \frac{\sin(x^2)}{x^2} = 0. \]