determine the behaviour of the function near the vertical asymptote of the following function: f(x) = 1/(2x…

determine the behaviour of the function near the vertical asymptote of the following function: f(x) = 1/(2x - 5) x→5/2⁻,y→ - ∞.x→5/2⁺,y→ - ∞. x→5/2⁻,y→∞.x→5/2⁺,y→ - ∞. x→5/2⁻,y→ - ∞.x→5/2⁺,y→∞. x→5/2⁻,y→∞.x→5/2⁺,y→∞

determine the behaviour of the function near the vertical asymptote of the following function: f(x) = 1/(2x - 5) x→5/2⁻,y→ - ∞.x→5/2⁺,y→ - ∞. x→5/2⁻,y→∞.x→5/2⁺,y→ - ∞. x→5/2⁻,y→ - ∞.x→5/2⁺,y→∞. x→5/2⁻,y→∞.x→5/2⁺,y→∞

Answer

Answer:

B. $x\to\frac{5}{2}^{-},y\to-\infty;x\to\frac{5}{2}^{+},y\to\infty$

Explanation:

Step1: Find vertical asymptote

Set denominator $2x - 5=0$, solve for $x$, we get $x=\frac{5}{2}$.

Step2: Analyze left - hand limit

Let $x\to\frac{5}{2}^{-}$, then $2x-5\to0^{-}$. So $f(x)=\frac{1}{2x - 5}\to-\infty$.

Step3: Analyze right - hand limit

Let $x\to\frac{5}{2}^{+}$, then $2x - 5\to0^{+}$. So $f(x)=\frac{1}{2x - 5}\to\infty$.