How does 41(s-1)/e^-s=H=lims→−∞ =-41/e^-s. What happen to the s before -1? Should it be 41s/e^-s
-41/e^-s?
To understand why the expression simplifies to -41/e^-s, let's break it down step by step.
We start with the expression 41(s-1)/e^-s and want to evaluate its limit as s approaches -∞.
1. First, let's factor out the common term e^-s from the denominator:
41(s-1)/(1/e^s)
2. When we flip the fraction in the denominator (1/x becomes x), we have:
(41(s-1)) * e^s
3. Now, let's evaluate the limit as s approaches -∞. As s becomes infinitely negative, e^s approaches 0.
So, we can rewrite the expression as:
(41(s-1)) * 0
4. Any number multiplied by 0 is equal to 0:
0 * (s-1) = 0
5. Therefore, the limit of 41(s-1)/e^-s as s approaches -∞ is 0.
Now, let's address the second part of your question. If we were to rewrite the expression as 41s/e^-s - 41/e^-s, we would end up with two terms. The first term, 41s/e^-s, would approach -∞ as s approaches -∞, while the second term, -41/e^-s, would approach 0. Therefore, the overall expression would not simplify to -41/e^-s because the first term would dominate the result and tend towards negative infinity.
In conclusion, the original expression simplifies to 0 as s approaches -∞, not -41/e^-s.