Questions: f(x)=(5x+1)/(x-1) lim x→-∞ f(x) =□ lim x→1⁻ f(x) =□ lim x→1⁺ f(x) =□ lim x→∞ f(x) =□

Solution Steps

To find the limit as \(x\) approaches negative infinity, we can analyze the behavior of the function for large negative values of \(x\). \[ \lim _{x \rightarrow-\infty} f(x)=\lim _{x \rightarrow-\infty} \frac{5 x+1}{x-1} \] We can divide both the numerator and the denominator by \(x\): \[ \lim _{x \rightarrow-\infty} \frac{5 x+1}{x-1}=\lim _{x \rightarrow-\infty} \frac{5+\frac{1}{x}}{1-\frac{1}{x}} \] As \(x\) approaches negative infinity, \(\frac{1}{x}\) approaches 0. Therefore, \[ \lim _{x \rightarrow-\infty} \frac{5+\frac{1}{x}}{1-\frac{1}{x}}=\frac{5+0}{1-0}=\frac{5}{1}=5 \]

To find the limit as \(x\) approaches 1 from the left (\(x \rightarrow 1^{-}\)), we consider values of \(x\) slightly less than 1. For example, if \(x = 0.9\), then \(5x + 1 = 5.5\) and \(x - 1 = -0.1\), so \(\frac{5x+1}{x-1} = \frac{5.5}{-0.1} = -55\). As \(x\) gets closer to 1 from the left, the numerator approaches 6, while the denominator approaches 0 from the negative side. Thus, the limit is negative infinity. \[ \lim _{x \rightarrow 1^{-}} f(x)=-\infty \]

To find the limit as \(x\) approaches 1 from the right (\(x \rightarrow 1^{+}\)), we consider values of \(x\) slightly greater than 1. For example, if \(x = 1.1\), then \(5x + 1 = 6.5\) and \(x - 1 = 0.1\), so \(\frac{5x+1}{x-1} = \frac{6.5}{0.1} = 65\). As \(x\) gets closer to 1 from the right, the numerator approaches 6, while the denominator approaches 0 from the positive side. Thus, the limit is positive infinity. \[ \lim _{x \rightarrow 1^{+}} f(x)=\infty \]

Final Answer