To determine the limit of the sequence
[tex]\[ d_n = \ln(n^2 + 5) - \ln(n^2 - 1) \][/tex]
as [tex]\( n \)[/tex] approaches infinity, follow these steps:
1. Apply properties of logarithms:
Recall the property of logarithms that states [tex]\( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \)[/tex]. Using this property, we can rewrite the sequence as:
[tex]\[ d_n = \ln\left(\frac{n^2 + 5}{n^2 - 1}\right) \][/tex]
2. Analyze the expression inside the logarithm:
To gain insight into the behavior of the sequence as [tex]\( n \)[/tex] approaches infinity, consider the fraction inside the logarithm:
[tex]\[ \frac{n^2 + 5}{n^2 - 1} \][/tex]
3. Simplify the fraction:
Divide both the numerator and the denominator by [tex]\( n^2 \)[/tex]:
[tex]\[ \frac{n^2 + 5}{n^2 - 1} = \frac{n^2/n^2 + 5/n^2}{n^2/n^2 - 1/n^2} = \frac{1 + \frac{5}{n^2}}{1 - \frac{1}{n^2}} \][/tex]
4. Take the limit as [tex]\( n \)[/tex] approaches infinity:
Observe that as [tex]\( n \)[/tex] grows larger and larger, the terms [tex]\( \frac{5}{n^2} \)[/tex] and [tex]\( \frac{1}{n^2} \)[/tex] approach zero:
[tex]\[
\lim_{n \rightarrow \infty} \frac{1 + \frac{5}{n^2}}{1 - \frac{1}{n^2}} = \frac{1 + 0}{1 - 0} = 1
\][/tex]
5. Apply the limit to the logarithm:
Now, we need to find the limit of the logarithm of this fraction as [tex]\( n \)[/tex] approaches infinity:
[tex]\[
\lim_{n \rightarrow \infty} \ln\left(\frac{n^2 + 5}{n^2 - 1}\right) = \ln\left(\lim_{n \rightarrow \infty} \frac{n^2 + 5}{n^2 - 1}\right) = \ln(1)
\][/tex]
6. Determine the final limit:
We know that [tex]\( \ln(1) = 0 \)[/tex].
Therefore, the limit of the sequence [tex]\( d_n \)[/tex] as [tex]\( n \)[/tex] approaches infinity is:
[tex]\[
\lim_{n \rightarrow \infty} d_n = 0
\][/tex]
Thus, the answer is [tex]\( \boxed{0} \)[/tex].
