Question

lim_{x\to\infty}\frac{\sqrt{x^{2}+11x}}{15 - 13x}

Explanation:

Step1: Divide numerator and denominator by \(x\)

When \(x\to\infty\), we have \(\lim_{x\to\infty}\frac{\sqrt{x^{2}+11x}}{15 - 13x}=\lim_{x\to\infty}\frac{\sqrt{x^{2}(1+\frac{11}{x})}}{x(\frac{15}{x}-13)}\). Since \(x>0\) as \(x\to\infty\), \(\sqrt{x^{2}} = x\), so it becomes \(\lim_{x\to\infty}\frac{x\sqrt{1 + \frac{11}{x}}}{x(\frac{15}{x}-13)}\).

Step2: Cancel out \(x\) and find the limit

Cancel out the non - zero factor \(x\) in the numerator and denominator, we get \(\lim_{x\to\infty}\frac{\sqrt{1+\frac{11}{x}}}{\frac{15}{x}-13}\). As \(x\to\infty\), \(\lim_{x\to\infty}\frac{11}{x}=0\) and \(\lim_{x\to\infty}\frac{15}{x}=0\). Then \(\lim_{x\to\infty}\frac{\sqrt{1+\frac{11}{x}}}{\frac{15}{x}-13}=\frac{\sqrt{1 + 0}}{0 - 13}=-\frac{1}{13}\).

Answer:

\(-\frac{1}{13}\)