1 Points] DETAILS MY NOTES SCALCET9 2.XP.6.006. Find the limit. (If the limit is infinite, enter ' \( \infty \) ' or '- \( \infty \) ', as appropriate. If the limit does not otherwise exist, enter ON \( \lim _{x \rightarrow \infty}\left(\frac{x-4}{x^{2}+5}\right) \)
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To find the limit as \( x \) approaches infinity for the expression \( \frac{x-4}{x^{2}+5} \), we can start by determining the highest power of \( x \) in both the numerator and denominator. The highest power in the numerator is \( x^1 \) (from \( x-4 \)) and in the denominator, it's \( x^2 \) (from \( x^2+5 \)). Now we can simplify the limit: \[ \lim_{x \to \infty} \frac{x-4}{x^2+5} = \lim_{x \to \infty} \frac{x(1 - \frac{4}{x})}{x^2(1 + \frac{5}{x^2})} \] Dividing both the numerator and the denominator by \( x^2 \): \[ = \lim_{x \to \infty} \frac{\frac{1 - \frac{4}{x}}{1 + \frac{5}{x^2}}} \] As \( x \) approaches infinity, the terms \( \frac{4}{x} \) and \( \frac{5}{x^2} \) both approach 0. Therefore, we have: \[ = \frac{1 - 0}{1 + 0} = \frac{1}{1} = 1 \] Thus, the limit is \[ 1 \]
