Question

limit laws and continuity: problem 1 (1 point) evaluate the limit (limlimits_{y \to 2} \frac{3(y^2 - 1)}{4y^2(y - 1)^3}) if the limit does not exist enter dne. limit = (square)

Explanation:

Step1: Factor the numerator

We know that \(y^2 - 1\) can be factored as a difference of squares: \(y^2 - 1=(y - 1)(y + 1)\). So the function becomes \(\frac{3(y - 1)(y + 1)}{4y^2(y - 1)^3}\).

Step2: Simplify the function

Cancel out the common factor \((y - 1)\) from the numerator and the denominator (note that \(y
eq1\) when taking the limit as \(y
ightarrow2\), so this cancellation is valid). We get \(\frac{3(y + 1)}{4y^2(y - 1)^2}\).

Step3: Substitute \(y = 2\) into the simplified function

Now we substitute \(y = 2\) into \(\frac{3(y + 1)}{4y^2(y - 1)^2}\).
First, calculate the numerator: \(3(2 + 1)=3\times3 = 9\).
Then, calculate the denominator: \(4\times(2)^2\times(2 - 1)^2=4\times4\times1 = 16\).

Answer:

\(\frac{9}{16}\)