Evaluate the limit \( \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8} \)
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To evaluate the limit \[ \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8}, \] we can first identify the highest power of \(x\) in both the numerator and the denominator. The numerator has a highest power of \(x^2\) and the denominator a highest power of \(x^1\). To simplify, we can divide every term in the numerator and denominator by \(x^2\): \[ \lim _{x \rightarrow \infty} \frac{11 - \frac{2}{x} + \frac{6}{x^2}}{\frac{4}{x} + \frac{8}{x^2}}. \] As \(x\) approaches infinity, the terms \(\frac{2}{x}\), \(\frac{6}{x^2}\), \(\frac{4}{x}\), and \(\frac{8}{x^2}\) will approach zero. Thus, we can simplify the limit to: \[ \lim _{x \rightarrow \infty} \frac{11 - 0 + 0}{0 + 0} = \frac{11}{0}. \] Since the degree of the numerator is higher than that of the denominator, the limit approaches infinity. Therefore, we conclude that: \[ \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8} = \infty. \]