2.3 the limits laws - l2: problem 3 (5 points) results for this submission the answer is not correct…
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.
Answer
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\rightarrow1}\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\rightarrow1}\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\rightarrow1), the limit is (\infty). But if we consider the one - sided limits: As (y\rightarrow1^{+}), (\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}\rightarrow+\infty) and as (y\rightarrow1^{-}), (\frac{6(y + 1)}{4y^{2}(y - 1)^{2}}\rightarrow+\infty). So the limit does not exist in the real - number system, and we write DNE.
Answer:
DNE