Question

2.3 the limits laws - l2: problem 3 (5 points) results for this submission the answer is not correct. evaluate the limit $lim_{y \to 1}\frac{6(y^{2}-1)}{4y^{2}(y - 1)^{3}}$ if the limit does not exist enter dne. limit = preview my answers submit answers your score was recorded. scores are sent to d2l brightspace every 24 hours. you have attempted this problem 4 times. you received a score of 0% for this attempt. your overall recorded score is 0%. you have 1 attempt remaining.

Explanation:

Step1: Factor the numerator

Use the difference - of - squares formula \(a^{2}-b^{2}=(a + b)(a - b)\). Here, \(y^{2}-1=(y + 1)(y - 1)\), so the function becomes \(\lim_{y
ightarrow1}\frac{6(y + 1)(y - 1)}{4y^{2}(y - 1)^{3}}\).

Step2: Simplify the function

Cancel out the common factor \((y - 1)\) in the numerator and denominator. We get \(\lim_{y
ightarrow1}\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}\).

Step3: Substitute \(y = 1\)

Substitute \(y = 1\) into the simplified function \(\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}\). The denominator \(4y^{2}(y - 1)^{2}=4\times1^{2}\times(1 - 1)^{2}=0\), and the numerator \(6(y + 1)=6\times(1 + 1)=12\). Since the numerator is non - zero and the denominator approaches 0 as \(y
ightarrow1\), the limit is \(\infty\). But if we consider the one - sided limits:
As \(y
ightarrow1^{+}\), \(\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}
ightarrow+\infty\) and as \(y
ightarrow1^{-}\), \(\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}
ightarrow+\infty\). So the limit does not exist in the real - number system, and we write DNE.

Answer:

DNE