Questions: Let f(x)=(2x+1)/(x^2-x-20) Determine each of the following values. If an answer does not exist, enter DNE. lim x→6^- f(x)=□

Transcript text: Let $f(x)=\frac{2 x+1}{x^{2}-x-20}$ Determine each of the following values. If an answer does not exist, enter DNE. \[ \lim _{x \rightarrow 6^{-}} f(x)=\square \]

Solution

Solution Steps

To find the limit of the function $ f(x) = \frac{2x+1}{x^2-x-20} $ as $ x $ approaches 6 from the left, we first need to factor the denominator to identify any potential discontinuities or vertical asymptotes. Once the function is factored, we can evaluate the behavior of the function as $ x $ approaches 6 from the left side.

Step 1: Factor the Denominator

The function given is $ f(x) = \frac{2x+1}{x^2-x-20} $. To find the limit as $ x $ approaches 6 from the left, we first factor the denominator. The quadratic expression $ x^2 - x - 20 $ can be factored as $ (x - 5)(x + 4) $.

Step 2: Identify Discontinuities

The factored form of the denominator, $ (x - 5)(x + 4) $, indicates potential vertical asymptotes at $ x = 5 $ and $ x = -4 $. Since we are interested in the limit as $ x \to 6^- $, these discontinuities do not directly affect our calculation.

Step 3: Evaluate the Limit

To find $ \lim_{x \to 6^-} f(x) $, we substitute values approaching 6 from the left into the function. Since 6 is not a point of discontinuity for the function, we can directly evaluate the limit:

\[ \lim_{x \to 6^-} \frac{2x+1}{(x-5)(x+4)} = \frac{2(6)+1}{(6-5)(6+4)} = \frac{13}{10} = 1.3 \]